Recall from Substitution Rule the method of integration by substitution. When evaluating an integral such as [latex]\displaystyle\int _2^3 x\left(x^{2}-4\right)^{5}dx[/latex], we substitute [latex]u=g(x)=x^{2}-4[/latex]. Then [latex]du=2x{dx}[/latex] or [latex]x{dx}=\frac{1}{2}{du}[/latex] and the limits change to [latex]u=g(2)=2^{2}-4=0[/latex] and [latex]u=g(3)=9-4=5[/latex]. Thus the integral becomes [latex]\displaystyle\int_0^5 \frac{1}{2}u^{5}du[/latex] and this integral is much simpler to evaluate. In other words, when solving integration problems, we make appropriate substitutions to obtain an integral that becomes much simpler than the original integral.
We also used this idea when we transformed double integrals in rectangular coordinates to polar coordinates and transformed triple integrals in rectangular coordinates to cylindrical or spherical coordinates to make the computations simpler. More generally,
[latex]\displaystyle\int_a^b f(x)dx= \displaystyle\int_c^d f(g(u))g^{\prime}(u)du[/latex],
Where [latex]x=g(u)[/latex], [latex]dx=g^{\prime}(u){du}[/latex], and [latex]u=c[/latex] and [latex]u=d[/latex] satisfy [latex]c=g(a)[/latex] and [latex]d=g(b)[/latex].
A similar result occurs in double integrals when we substitute [latex]x=h(r,\theta)=r\cos{\theta}[/latex], [latex]y=g(r,\theta)=r\sin{\theta}[/latex], and [latex]dA=dxdy=rdrd\theta[/latex]. Then we get
[latex]\underset{R}{\displaystyle\iint} f(x,y)dA= \underset{S}{\displaystyle\iint} f(r\cos{\theta},r\sin{\theta})rdrd\theta[/latex]
where the domain [latex]R[/latex] is replaced by the domain [latex]S[/latex] in polar coordinates. Generally, the function that we use to change the variables to make the integration simpler is called a transformation or mapping.
Candela Citations
- Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/books/calculus-volume-3/pages/1-introduction. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction