Learning Outcomes
- Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals
- Use a table of integrals to solve integration problems
In the Arc Length and Curvature section, we will explore the curvature of vector-valued functions. Here we will review how to apply the Second Fundamental Theorem of Calculus and use integration table formulas.
Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem
The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.
The Fundamental Theorem of Calculus, Part 2
If [latex]f[/latex] is continuous over the interval [latex]\left[a,b\right][/latex] and [latex]F(x)[/latex] is any antiderivative of [latex]f(x),[/latex] then
[latex]{\displaystyle\int }_{a}^{b}f(x)dx=F(b)-F(a)[/latex]
Example: Evaluating an Integral with the Fundamental Theorem of Calculus
Use the second part of the Fundamental Theorem of Calculus to evaluate
[latex]{\displaystyle\int }_{-2}^{2}({t}^{2}-4)dt.[/latex]
Show Solution
Recall the power rule for Antiderivatives:
[latex]\text{ If }y={x}^{n},\displaystyle\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C.[/latex]
Use this rule to find the antiderivative of the function and then apply the theorem. We have
[latex]\begin{array}{cc}{\displaystyle\int }_{-2}^{2}({t}^{2}-4)dt\hfill & =\frac{{t}^{3}}{3}-{4t|}_{-2}^{2}\hfill \\ \\ \\ & =\left[\frac{{(2)}^{3}}{3}-4(2)\right]-\left[\frac{{(-2)}^{3}}{3}-4(-2)\right]\hfill \\ & =(\frac{8}{3}-8)-(-\frac{8}{3}+8)\hfill \\ & =\frac{8}{3}-8+\frac{8}{3}-8\hfill \\ & =\frac{16}{3}-16\hfill \\ & =-\frac{32}{3}.\hfill \end{array}[/latex]
Example: Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:
[latex]{\displaystyle\int }_{1}^{9}\dfrac{x-1}{\sqrt{x}}dx.[/latex]
Show Solution
First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:
[latex]{\displaystyle\int }_{1}^{9}\frac{x-1}{{x}^{1\text{/}2}}dx={\displaystyle\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx\text{.}[/latex]
Use the properties of exponents to simplify:
[latex]{\displaystyle\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx={\displaystyle\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx\text{.}[/latex]
Now, integrate using the power rule:
[latex]\begin{array}{}\\ \\ {\displaystyle\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx\hfill & ={(\frac{{x}^{3\text{/}2}}{\frac{3}{2}}-\frac{{x}^{1\text{/}2}}{\frac{1}{2}})|}_{1}^{9}\hfill \\ \\ & =\left[\frac{{(9)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(9)}^{1\text{/}2}}{\frac{1}{2}}\right]-\left[\frac{{(1)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(1)}^{1\text{/}2}}{\frac{1}{2}}\right]\hfill \\ & =\left[\frac{2}{3}(27)-2(3)\right]-\left[\frac{2}{3}(1)-2(1)\right]\hfill \\ & =18-6-\frac{2}{3}+2\hfill \\ & =\frac{40}{3}.\hfill \end{array}[/latex]
Try It
Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\displaystyle\int }_{1}^{2}{x}^{-4}dx.[/latex]
Show Solution
[latex]\frac{7}{24}[/latex]
Tables of Integrals
Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution [latex]x=\tan\theta [/latex], we can arrive at
[latex]\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)+C[/latex].
However, using [latex]x=\text{sinh}\theta [/latex], we obtained a different solution—namely,
[latex]\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}={\text{sinh}}^{-1}x+C[/latex].
We later showed algebraically that the two solutions are equivalent. That is, we showed that [latex]{\text{sinh}}^{-1}x=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)[/latex]. In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as the difference in the two antiderivatives is a constant, they are equivalent.
Example: Using a Formula from a Table to Evaluate an Integral
Use the table formula
[latex]\displaystyle\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\sin}^{-1}\frac{u}{a}+C[/latex]
to evaluate [latex]\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[/latex].
Show Solution
If we look at integration tables, we see that several formulas contain expressions of the form [latex]\sqrt{{a}^{2}-{u}^{2}}[/latex]. This expression is actually similar to [latex]\sqrt{16-{e}^{2x}}[/latex], where [latex]a=4[/latex] and [latex]u={e}^{x}[/latex]. Keep in mind that we must also have [latex]du={e}^{x}[/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[/latex] should help to put this integral in a useful form. Thus, we now have
[latex]\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx[/latex].
Substituting [latex]u={e}^{x}[/latex] and [latex]du={e}^{x}[/latex] produces [latex]\displaystyle\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[/latex]. From the integration table (#88 in Appendix A),
[latex]\displaystyle\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\sin}^{-1}\frac{u}{a}+C[/latex].
Thus,
[latex]\begin{array}{ccccc}\hfill {\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx}& ={\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx}\hfill & & & \text{Substitute}u={e}^{x}\text{and}du={e}^{x}dx.\hfill \\ & ={\displaystyle\int \frac{\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du}\hfill & & & \text{Apply the formula using}a=4.\hfill \\ & =-\frac{\sqrt{{4}^{2}-{u}^{2}}}{u}-{\sin}^{-1}\frac{u}{4}+C\hfill & & & \text{Substitute}u={e}^{x}.\hfill \\ & =-\frac{\sqrt{16-{e}^{2x}}}{u}-{\sin}^{-1}\left(\frac{{e}^{x}}{4}\right)+C.\hfill & & & \end{array}[/latex]
The full integration table with formulas can be found in Appendix A.
