{"id":1156,"date":"2021-06-30T17:02:01","date_gmt":"2021-06-30T17:02:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-substitution\/"},"modified":"2021-12-09T00:05:19","modified_gmt":"2021-12-09T00:05:19","slug":"problem-set-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-substitution\/","title":{"raw":"Problem Set: Substitution","rendered":"Problem Set: Substitution"},"content":{"raw":"
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1. <\/strong>Why is [latex]u[\/latex]-substitution referred to as change of variable<\/em>?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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2.<\/strong> If [latex]f=g\\circ h,[\/latex] when reversing the chain rule, [latex]\\frac{d}{dx}(g\\circ h)(x)={g}^{\\prime }(h(x)){h}^{\\prime }(x),[\/latex] should you take [latex]u=g(x)[\/latex] or [latex]u=h(x)?[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587726\"]\r\n

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[latex]u=h(x)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, verify each identity using differentiation. Then, using the indicated [latex]u[\/latex]-substitution, identify [latex]f[\/latex] such that the integral takes the form [latex]\\displaystyle\\int f(u)du.[\/latex]<\/p>\r\n\r\n

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3. <\/strong>[latex]\\displaystyle\\int x\\sqrt{x+1}dx=\\frac{2}{15}{(x+1)}^{3\\text{\/}2}(3x-2)+C; \\,\\,\\, u=x+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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4. <\/strong>For [latex]x>1[\/latex]:\u00a0 [latex]\\displaystyle\\int \\frac{{x}^{2}}{\\sqrt{x-1}}dx = \\frac{2}{15}\\sqrt{x-1}(3{x}^{2}+4x+8)+C; \\,\\,\\, u=x-1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587727\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587727\"]\r\n

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[latex]f(u)=\\frac{{(u+1)}^{2}}{\\sqrt{u}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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5. <\/strong>[latex]\\displaystyle\\int x\\sqrt{4{x}^{2}+9}dx=\\frac{1}{12}{(4{x}^{2}+9)}^{3\\text{\/}2}+C; \\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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6. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{4{x}^{2}+9}}dx=\\frac{1}{4}\\sqrt{4{x}^{2}+9}+C;\\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587728\"]\r\n

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[latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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7. <\/strong>[latex]\\displaystyle\\int \\frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\\frac{1}{8(4{x}^{2}+9)};\\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the antiderivative using the indicated substitution.<\/p>\r\n\r\n

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8. <\/strong>[latex]\\displaystyle\\int {(x+1)}^{4}dx; \\,\\,\\, u=x+1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587729\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587729\"]\r\n

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[latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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9. <\/strong>[latex]\\displaystyle\\int {(x-1)}^{5}dx; \\,\\,\\, u=x-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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10. <\/strong>[latex]\\displaystyle\\int {(2x-3)}^{-7}dx; \\,\\,\\, u=2x-3[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587730\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587730\"]\r\n

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[latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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11. <\/strong>[latex]\\displaystyle\\int {(3x-2)}^{-11}dx; \\,\\,\\, u=3x-2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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12. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{{x}^{2}+1}}dx; \\,\\,\\, u={x}^{2}+1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587731\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587731\"]\r\n

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[latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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13. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{1-{x}^{2}}}dx; \\,\\,\\, u=1-{x}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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14. <\/strong>[latex]\\displaystyle\\int (x-1){({x}^{2}-2x)}^{3}dx; \\,\\,\\, u={x}^{2}-2x[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587732\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587732\"]\r\n

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[latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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15. <\/strong>[latex]\\displaystyle\\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx; \\,\\,\\, u={x}^{3}-3{x}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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16. <\/strong>[latex]\\displaystyle\\int { \\cos }^{3}\\theta d\\theta ; \\,\\,\\, u= \\sin \\theta [\/latex]<\/p>\r\n[reveal-answer q=\"41198867\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"41198867\"]\r\n\r\n([latex]\\cos ^{2}\\theta =1-{ \\sin }^{2}\\theta [\/latex])\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572587733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587733\"]\r\n

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[latex] \\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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17. <\/strong>[latex]\\displaystyle\\int { \\sin }^{3}\\theta d\\theta ; \\,\\,\\, u= \\cos \\theta [\/latex]<\/p>\r\n[reveal-answer q=\"77553442\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"77553442\"]\r\n\r\n[latex] \\sin ^{2}\\theta =1-{ \\cos }^{2}\\theta[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, use a suitable change of variables to determine the indefinite integral.<\/p>\r\n\r\n

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18. <\/strong>[latex]\\displaystyle\\int x{(1-x)}^{99}dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587734\"]\r\n

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[latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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19. <\/strong>[latex]\\displaystyle\\int t{(1-{t}^{2})}^{10}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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20. <\/strong>[latex]\\displaystyle\\int {(11x-7)}^{-3}dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587735\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587735\"]\r\n

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[latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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21. <\/strong>[latex]\\displaystyle\\int {(7x-11)}^{4}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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22. <\/strong>[latex]\\displaystyle\\int { \\cos }^{3}\\theta \\sin \\theta d\\theta [\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587736\"]\r\n

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[latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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23. <\/strong>[latex]\\displaystyle\\int { \\sin }^{7}\\theta \\cos \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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24. <\/strong>[latex]\\displaystyle\\int { \\cos }^{2}(\\pi t) \\sin (\\pi t)dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587737\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587737\"]\r\n

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[latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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25. <\/strong>[latex]\\displaystyle\\int { \\sin }^{2}x{ \\cos }^{3}xdx[\/latex]<\/p>\r\n[reveal-answer q=\"68800932\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"68800932\"]\r\n\r\n[latex]\\sin ^{2}x+{ \\cos }^{2}x=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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26. <\/strong>[latex]\\displaystyle\\int [latex]t[\/latex] \\sin ({t}^{2}) \\cos ({t}^{2})dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587738\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587738\"]\r\n

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[latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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27. <\/strong>[latex]\\displaystyle\\int {t}^{2}{ \\cos }^{2}({t}^{3}) \\sin ({t}^{3})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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28. <\/strong>[latex]\\displaystyle\\int \\frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587739\"]\r\n

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[latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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29. <\/strong>[latex]\\displaystyle\\int \\frac{{x}^{3}}{\\sqrt{1-{x}^{2}}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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30. <\/strong>[latex]\\displaystyle\\int \\frac{{y}^{5}}{{(1-{y}^{3})}^{3\\text{\/}2}}dy[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587740\"]\r\n

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[latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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31. <\/strong>[latex]{\\displaystyle\\int \\cos \\theta (1- \\cos \\theta )}^{99} \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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32. <\/strong>[latex]{\\displaystyle\\int (1-{ \\cos }^{3}\\theta )}^{10}{ \\cos }^{2}\\theta \\sin \\theta d\\theta [\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587741\"]\r\n

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[latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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33. <\/strong>[latex]\\displaystyle\\int ( \\cos \\theta -1){({ \\cos }^{2}\\theta -2 \\cos \\theta )}^{3} \\sin \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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34. <\/strong>[latex]\\displaystyle\\int ({ \\sin }^{2}\\theta -2 \\sin \\theta ){({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{3} \\cos \\theta d\\theta [\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587742\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587742\"]\r\n

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[latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nIn the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.\r\n

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35. [T]<\/strong> [latex]y=3{(1-x)}^{2}[\/latex] over [latex]\\left[0,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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36. [T]<\/strong> [latex]y=x{(1-{x}^{2})}^{3}[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587743\"]\r\n

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[latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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37. [T]<\/strong> [latex]y= \\sin x{(1- \\cos x)}^{2}[\/latex] over [latex]\\left[0,\\pi \\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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38. [T]<\/strong> [latex]y=\\dfrac{x}{{({x}^{2}+1)}^{2}}[\/latex] over [latex]\\left[-1,1\\right][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587744\"]\r\n

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[latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is 0.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, use a change of variables to evaluate the definite integral.<\/p>\r\n\r\n

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39. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}x\\sqrt{1-{x}^{2}}dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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40. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{x}{\\sqrt{1+{x}^{2}}}dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587745\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587745\"]\r\n

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[latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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41. <\/strong>[latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{t^2}{\\sqrt{5+{t}^{2}}}dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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42. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{t^2}{\\sqrt{1+{t}^{3}}}dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587746\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587746\"]\r\n

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[latex]u=1+{t}^{3},du=3{t}^{2}du,\\frac{1}{3}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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43. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}{ \\sec }^{2}\\theta \\tan \\theta d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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44. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}\\dfrac{ \\sin \\theta }{{ \\cos }^{4}\\theta }d\\theta [\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587747\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587747\"]\r\n

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[latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\displaystyle\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, evaluate the indefinite integral [latex]\\displaystyle\\int f(x)dx[\/latex] with constant [latex]C=0[\/latex] using [latex]u[\/latex]-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C<\/em> that would need to be added to the antiderivative to make it equal to the definite integral [latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex] with [latex]a[\/latex] the left endpoint of the given interval.<\/p>\r\n\r\n

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45. [T]<\/strong> [latex]\\displaystyle\\int (2x+1){e}^{{x}^{2}+x-6}dx[\/latex] over [latex]\\left[-3,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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46. [T]<\/strong> [latex]\\displaystyle\\int \\frac{ \\cos (\\text{ln}(2x))}{x}dx[\/latex] on [latex]\\left[0,2\\right][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587748\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587748\"]\r\n

\"Two<\/span>\r\nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.\u00a0 [\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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47. [T]<\/strong> [latex]\\displaystyle\\int \\frac{3{x}^{2}+2x+1}{\\sqrt{{x}^{3}+{x}^{2}+x+4}}dx[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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48. [T]<\/strong> [latex]\\displaystyle\\int \\frac{ \\sin x}{{ \\cos }^{3}x}dx[\/latex] over [latex]\\left[-\\frac{\\pi }{3},\\frac{\\pi }{3}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587749\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587749\"]\r\n

\"Two<\/span>\r\nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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49. [T]<\/strong> [latex]\\displaystyle\\int (x+2){e}^{\\text{\u2212}{x}^{2}-4x+3}dx[\/latex] over [latex]\\left[-5,1\\right][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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50. [T]<\/strong> [latex]\\displaystyle\\int 3{x}^{2}\\sqrt{2{x}^{3}+1}dx[\/latex] over [latex]\\left[0,1\\right][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587750\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587750\"]\r\n

\"Two<\/span>\r\nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex] [\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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51. <\/strong>If [latex]h(a)=h(b)[\/latex] in [latex]{\\displaystyle\\int }_{a}^{b}g\\text{\u2018}(h(x))h(x)dx,[\/latex] what can you say about the value of the integral?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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52. <\/strong>Is the substitution [latex]u=1-{x}^{2}[\/latex] in the definite integral [latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{x}{1-{x}^{2}}dx[\/latex] okay? If not, why not?<\/p>\r\n[reveal-answer q=\"fs-id1170572587752\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587752\"]\r\n

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No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, use a change of variables to show that each definite integral is equal to zero.<\/p>\r\n\r\n

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53. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi }{ \\cos }^{2}(2\\theta ) \\sin (2\\theta )d\\theta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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54. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\sqrt{\\pi }}t \\cos ({t}^{2}) \\sin ({t}^{2})dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587754\"]\r\n

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[latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{0}udu.[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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55. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}(1-2t)dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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56. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{1-2t}{(1+{(t-\\frac{1}{2})}^{2})}dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587756\"]\r\n

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[latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\displaystyle\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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57. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi } \\sin ({(t-\\frac{\\pi }{2})}^{3}) \\cos (t-\\frac{\\pi }{2})dt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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58. <\/strong>[latex]{\\displaystyle\\int }_{0}^{2}(1-t) \\cos (\\pi t)dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587758\"]\r\n

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[latex]u=1-t;[\/latex] the integral becomes<\/p>\r\n[latex]\\begin{array}{l}{\\displaystyle\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\displaystyle\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\displaystyle\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\displaystyle\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]\r\nsince the integrand is odd.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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59. <\/strong>[latex]{\\displaystyle\\int }_{\\pi \\text{\/}4}^{3\\pi \\text{\/}4}{ \\sin }^{2}t \\cos tdt[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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60. <\/strong>Show that the average value of [latex]f(x)[\/latex] over an interval [latex]\\left[a,b\\right][\/latex] is the same as the average value of [latex]f(cx)[\/latex] over the interval [latex]\\left[\\frac{a}{c},\\frac{b}{c}\\right][\/latex] for [latex]c>0.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587760\"]\r\n

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Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\displaystyle\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\displaystyle\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(u)du.[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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61. <\/strong>Find the area under the graph of [latex]f(t)=\\dfrac{t}{{(1+{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x[\/latex] where [latex]a>0[\/latex] and [latex]a\\ne 1[\/latex] is fixed, and evaluate the limit as [latex]x\\to \\infty .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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62. <\/strong>Find the area under the graph of [latex]g(t)=\\dfrac{t}{{(1-{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x,[\/latex] where [latex]0<x<1[\/latex] and [latex]a>0[\/latex] is fixed. Evaluate the limit as [latex]x\\to 1.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572587762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587762\"]\r\n

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[latex]{\\displaystyle\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\displaystyle\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a<1,[\/latex] and the limit diverges to +\u221e if [latex]a>1.[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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63. <\/strong>The area of a semicircle of radius 1 can be expressed as [latex]{\\displaystyle\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx.[\/latex] Use the substitution [latex]x= \\cos t[\/latex] to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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64. <\/strong>The area of the top half of an ellipse with a major axis that is the [latex]x[\/latex]-axis from [latex]x=-1[\/latex] to [latex]a[\/latex] and with a minor axis that is the [latex]y[\/latex]-axis from [latex]y=\\text{\u2212}b[\/latex] to [latex]b[\/latex] can be written as [latex]{\\displaystyle\\int }_{\\text{\u2212}a}^{a}b\\sqrt{1-\\frac{{x}^{2}}{{a}^{2}}}dx.[\/latex] Use the substitution [latex]x=a \\cos t[\/latex] to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.<\/p>\r\n[reveal-answer q=\"fs-id1170572587764\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587764\"]\r\n

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[latex]{\\displaystyle\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\displaystyle\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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65. [T]<\/strong> The following graph is of a function of the form [latex]f(t)=a \\sin (nt)+b \\sin (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex], and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\displaystyle\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\r\n\"A<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

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66. [T]<\/strong> The following graph is of a function of the form [latex]f(x)=a \\cos (nt)+b \\cos (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex] and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\displaystyle\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\r\n\"The<\/span>\r\n\r\n[reveal-answer q=\"fs-id1170572587766\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572587766\"]\r\n

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[latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

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1. <\/strong>Why is [latex]u[\/latex]-substitution referred to as change of variable<\/em>?<\/p>\n<\/div>\n<\/div>\n

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2.<\/strong> If [latex]f=g\\circ h,[\/latex] when reversing the chain rule, [latex]\\frac{d}{dx}(g\\circ h)(x)={g}^{\\prime }(h(x)){h}^{\\prime }(x),[\/latex] should you take [latex]u=g(x)[\/latex] or [latex]u=h(x)?[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]u=h(x)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, verify each identity using differentiation. Then, using the indicated [latex]u[\/latex]-substitution, identify [latex]f[\/latex] such that the integral takes the form [latex]\\displaystyle\\int f(u)du.[\/latex]<\/p>\n

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3. <\/strong>[latex]\\displaystyle\\int x\\sqrt{x+1}dx=\\frac{2}{15}{(x+1)}^{3\\text{\/}2}(3x-2)+C; \\,\\,\\, u=x+1[\/latex]<\/p>\n<\/div>\n<\/div>\n

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4. <\/strong>For [latex]x>1[\/latex]:\u00a0 [latex]\\displaystyle\\int \\frac{{x}^{2}}{\\sqrt{x-1}}dx = \\frac{2}{15}\\sqrt{x-1}(3{x}^{2}+4x+8)+C; \\,\\,\\, u=x-1[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]f(u)=\\frac{{(u+1)}^{2}}{\\sqrt{u}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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5. <\/strong>[latex]\\displaystyle\\int x\\sqrt{4{x}^{2}+9}dx=\\frac{1}{12}{(4{x}^{2}+9)}^{3\\text{\/}2}+C; \\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\n<\/div>\n<\/div>\n

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6. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{4{x}^{2}+9}}dx=\\frac{1}{4}\\sqrt{4{x}^{2}+9}+C;\\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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7. <\/strong>[latex]\\displaystyle\\int \\frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\\frac{1}{8(4{x}^{2}+9)};\\,\\,\\, u=4{x}^{2}+9[\/latex]<\/p>\n<\/div>\n<\/div>\n

In the following exercises, find the antiderivative using the indicated substitution.<\/p>\n

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8. <\/strong>[latex]\\displaystyle\\int {(x+1)}^{4}dx; \\,\\,\\, u=x+1[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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9. <\/strong>[latex]\\displaystyle\\int {(x-1)}^{5}dx; \\,\\,\\, u=x-1[\/latex]<\/p>\n<\/div>\n<\/div>\n

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10. <\/strong>[latex]\\displaystyle\\int {(2x-3)}^{-7}dx; \\,\\,\\, u=2x-3[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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11. <\/strong>[latex]\\displaystyle\\int {(3x-2)}^{-11}dx; \\,\\,\\, u=3x-2[\/latex]<\/p>\n<\/div>\n<\/div>\n

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12. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{{x}^{2}+1}}dx; \\,\\,\\, u={x}^{2}+1[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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13. <\/strong>[latex]\\displaystyle\\int \\frac{x}{\\sqrt{1-{x}^{2}}}dx; \\,\\,\\, u=1-{x}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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14. <\/strong>[latex]\\displaystyle\\int (x-1){({x}^{2}-2x)}^{3}dx; \\,\\,\\, u={x}^{2}-2x[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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15. <\/strong>[latex]\\displaystyle\\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx; \\,\\,\\, u={x}^{3}-3{x}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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16. <\/strong>[latex]\\displaystyle\\int { \\cos }^{3}\\theta d\\theta ; \\,\\,\\, u= \\sin \\theta[\/latex]<\/p>\n

Hint<\/span><\/p>\n
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([latex]\\cos ^{2}\\theta =1-{ \\sin }^{2}\\theta[\/latex])<\/p>\n<\/div>\n<\/div>\n

Show Solution<\/span><\/p>\n
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[latex]\\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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17. <\/strong>[latex]\\displaystyle\\int { \\sin }^{3}\\theta d\\theta ; \\,\\,\\, u= \\cos \\theta[\/latex]<\/p>\n

Hint<\/span><\/p>\n
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[latex]\\sin ^{2}\\theta =1-{ \\cos }^{2}\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, use a suitable change of variables to determine the indefinite integral.<\/p>\n

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18. <\/strong>[latex]\\displaystyle\\int x{(1-x)}^{99}dx[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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19. <\/strong>[latex]\\displaystyle\\int t{(1-{t}^{2})}^{10}dt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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20. <\/strong>[latex]\\displaystyle\\int {(11x-7)}^{-3}dx[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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21. <\/strong>[latex]\\displaystyle\\int {(7x-11)}^{4}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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22. <\/strong>[latex]\\displaystyle\\int { \\cos }^{3}\\theta \\sin \\theta d\\theta[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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23. <\/strong>[latex]\\displaystyle\\int { \\sin }^{7}\\theta \\cos \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n

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24. <\/strong>[latex]\\displaystyle\\int { \\cos }^{2}(\\pi t) \\sin (\\pi t)dt[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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25. <\/strong>[latex]\\displaystyle\\int { \\sin }^{2}x{ \\cos }^{3}xdx[\/latex]<\/p>\n

Hint<\/span><\/p>\n
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[latex]\\sin ^{2}x+{ \\cos }^{2}x=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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26. <\/strong>[latex]\\displaystyle\\int [latex]t[\/latex] \\sin ({t}^{2}) \\cos ({t}^{2})dt[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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27. <\/strong>[latex]\\displaystyle\\int {t}^{2}{ \\cos }^{2}({t}^{3}) \\sin ({t}^{3})dt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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28. <\/strong>[latex]\\displaystyle\\int \\frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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29. <\/strong>[latex]\\displaystyle\\int \\frac{{x}^{3}}{\\sqrt{1-{x}^{2}}}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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30. <\/strong>[latex]\\displaystyle\\int \\frac{{y}^{5}}{{(1-{y}^{3})}^{3\\text{\/}2}}dy[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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31. <\/strong>[latex]{\\displaystyle\\int \\cos \\theta (1- \\cos \\theta )}^{99} \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n

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32. <\/strong>[latex]{\\displaystyle\\int (1-{ \\cos }^{3}\\theta )}^{10}{ \\cos }^{2}\\theta \\sin \\theta d\\theta[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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33. <\/strong>[latex]\\displaystyle\\int ( \\cos \\theta -1){({ \\cos }^{2}\\theta -2 \\cos \\theta )}^{3} \\sin \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n

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34. <\/strong>[latex]\\displaystyle\\int ({ \\sin }^{2}\\theta -2 \\sin \\theta ){({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{3} \\cos \\theta d\\theta[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.<\/p>\n

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35. [T]<\/strong> [latex]y=3{(1-x)}^{2}[\/latex] over [latex]\\left[0,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

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36. [T]<\/strong> [latex]y=x{(1-{x}^{2})}^{3}[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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37. [T]<\/strong> [latex]y= \\sin x{(1- \\cos x)}^{2}[\/latex] over [latex]\\left[0,\\pi \\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

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38. [T]<\/strong> [latex]y=\\dfrac{x}{{({x}^{2}+1)}^{2}}[\/latex] over [latex]\\left[-1,1\\right][\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, use a change of variables to evaluate the definite integral.<\/p>\n

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39. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}x\\sqrt{1-{x}^{2}}dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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40. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{x}{\\sqrt{1+{x}^{2}}}dx[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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41. <\/strong>[latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{t^2}{\\sqrt{5+{t}^{2}}}dt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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42. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{t^2}{\\sqrt{1+{t}^{3}}}dt[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]u=1+{t}^{3},du=3{t}^{2}du,\\frac{1}{3}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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43. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}{ \\sec }^{2}\\theta \\tan \\theta d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n

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44. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}\\dfrac{ \\sin \\theta }{{ \\cos }^{4}\\theta }d\\theta[\/latex]<\/p>\n

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[latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\displaystyle\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, evaluate the indefinite integral [latex]\\displaystyle\\int f(x)dx[\/latex] with constant [latex]C=0[\/latex] using [latex]u[\/latex]-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C<\/em> that would need to be added to the antiderivative to make it equal to the definite integral [latex]F(x)={\\displaystyle\\int }_{a}^{x}f(t)dt,[\/latex] with [latex]a[\/latex] the left endpoint of the given interval.<\/p>\n

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45. [T]<\/strong> [latex]\\displaystyle\\int (2x+1){e}^{{x}^{2}+x-6}dx[\/latex] over [latex]\\left[-3,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

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46. [T]<\/strong> [latex]\\displaystyle\\int \\frac{ \\cos (\\text{ln}(2x))}{x}dx[\/latex] on [latex]\\left[0,2\\right][\/latex]<\/p>\n

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\nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.\u00a0 <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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47. [T]<\/strong> [latex]\\displaystyle\\int \\frac{3{x}^{2}+2x+1}{\\sqrt{{x}^{3}+{x}^{2}+x+4}}dx[\/latex] over [latex]\\left[-1,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

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48. [T]<\/strong> [latex]\\displaystyle\\int \\frac{ \\sin x}{{ \\cos }^{3}x}dx[\/latex] over [latex]\\left[-\\frac{\\pi }{3},\\frac{\\pi }{3}\\right][\/latex]<\/p>\n

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\"Two<\/span>
\nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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49. [T]<\/strong> [latex]\\displaystyle\\int (x+2){e}^{\\text{\u2212}{x}^{2}-4x+3}dx[\/latex] over [latex]\\left[-5,1\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n

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50. [T]<\/strong> [latex]\\displaystyle\\int 3{x}^{2}\\sqrt{2{x}^{3}+1}dx[\/latex] over [latex]\\left[0,1\\right][\/latex]<\/p>\n

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\"Two<\/span>
\nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex] <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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51. <\/strong>If [latex]h(a)=h(b)[\/latex] in [latex]{\\displaystyle\\int }_{a}^{b}g\\text{\u2018}(h(x))h(x)dx,[\/latex] what can you say about the value of the integral?<\/p>\n<\/div>\n<\/div>\n

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52. <\/strong>Is the substitution [latex]u=1-{x}^{2}[\/latex] in the definite integral [latex]{\\displaystyle\\int }_{0}^{2}\\dfrac{x}{1-{x}^{2}}dx[\/latex] okay? If not, why not?<\/p>\n

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No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

In the following exercises, use a change of variables to show that each definite integral is equal to zero.<\/p>\n

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53. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi }{ \\cos }^{2}(2\\theta ) \\sin (2\\theta )d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n

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54. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\sqrt{\\pi }}t \\cos ({t}^{2}) \\sin ({t}^{2})dt[\/latex]<\/p>\n

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[latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{0}udu.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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55. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}(1-2t)dt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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56. <\/strong>[latex]{\\displaystyle\\int }_{0}^{1}\\dfrac{1-2t}{(1+{(t-\\frac{1}{2})}^{2})}dt[\/latex]<\/p>\n

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[latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\displaystyle\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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57. <\/strong>[latex]{\\displaystyle\\int }_{0}^{\\pi } \\sin ({(t-\\frac{\\pi }{2})}^{3}) \\cos (t-\\frac{\\pi }{2})dt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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58. <\/strong>[latex]{\\displaystyle\\int }_{0}^{2}(1-t) \\cos (\\pi t)dt[\/latex]<\/p>\n

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[latex]u=1-t;[\/latex] the integral becomes<\/p>\n

[latex]\\begin{array}{l}{\\displaystyle\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\displaystyle\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\displaystyle\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\displaystyle\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]
\nsince the integrand is odd.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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59. <\/strong>[latex]{\\displaystyle\\int }_{\\pi \\text{\/}4}^{3\\pi \\text{\/}4}{ \\sin }^{2}t \\cos tdt[\/latex]<\/p>\n<\/div>\n<\/div>\n

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60. <\/strong>Show that the average value of [latex]f(x)[\/latex] over an interval [latex]\\left[a,b\\right][\/latex] is the same as the average value of [latex]f(cx)[\/latex] over the interval [latex]\\left[\\frac{a}{c},\\frac{b}{c}\\right][\/latex] for [latex]c>0.[\/latex]<\/p>\n

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Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\displaystyle\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\displaystyle\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(u)du.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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61. <\/strong>Find the area under the graph of [latex]f(t)=\\dfrac{t}{{(1+{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x[\/latex] where [latex]a>0[\/latex] and [latex]a\\ne 1[\/latex] is fixed, and evaluate the limit as [latex]x\\to \\infty .[\/latex]<\/p>\n<\/div>\n<\/div>\n

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62. <\/strong>Find the area under the graph of [latex]g(t)=\\dfrac{t}{{(1-{t}^{2})}^{a}}[\/latex] between [latex]t=0[\/latex] and [latex]t=x,[\/latex] where [latex]00[\/latex] is fixed. Evaluate the limit as [latex]x\\to 1.[\/latex]<\/p>\n

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[latex]{\\displaystyle\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\displaystyle\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a<1,[\/latex] and the limit diverges to +\u221e if [latex]a>1.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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63. <\/strong>The area of a semicircle of radius 1 can be expressed as [latex]{\\displaystyle\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx.[\/latex] Use the substitution [latex]x= \\cos t[\/latex] to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.<\/p>\n<\/div>\n<\/div>\n

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64. <\/strong>The area of the top half of an ellipse with a major axis that is the [latex]x[\/latex]-axis from [latex]x=-1[\/latex] to [latex]a[\/latex] and with a minor axis that is the [latex]y[\/latex]-axis from [latex]y=\\text{\u2212}b[\/latex] to [latex]b[\/latex] can be written as [latex]{\\displaystyle\\int }_{\\text{\u2212}a}^{a}b\\sqrt{1-\\frac{{x}^{2}}{{a}^{2}}}dx.[\/latex] Use the substitution [latex]x=a \\cos t[\/latex] to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.<\/p>\n

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[latex]{\\displaystyle\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\displaystyle\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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65. [T]<\/strong> The following graph is of a function of the form [latex]f(t)=a \\sin (nt)+b \\sin (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex], and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\displaystyle\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\n

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66. [T]<\/strong> The following graph is of a function of the form [latex]f(x)=a \\cos (nt)+b \\cos (mt).[\/latex] Estimate the coefficients [latex]a[\/latex] and [latex]b[\/latex] and the frequency parameters [latex]n[\/latex] and [latex]m[\/latex]. Use these estimates to approximate [latex]{\\displaystyle\\int }_{0}^{\\pi }f(t)dt.[\/latex]<\/p>\n

\"The<\/span><\/p>\n

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[latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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