Double Integration and Applications: Background You’ll Need 1
Evaluate definite integrals using integration techniques including substitution, and integration by parts
Substitution for Definite Integrals
Substitution can be used with definite integrals. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
substitution with definite integrals
Let [latex]u=g(x)[/latex] and let [latex]{g}^{\text{′}}[/latex] be continuous over an interval [latex]\left[a,b\right],[/latex] and let [latex]f[/latex] be continuous over the range of [latex]u=g(x).[/latex] Then,
Use substitution to evaluate [latex]{\displaystyle\int }_{0}^{1}{x}^{2}{(1+2{x}^{3})}^{5}dx.[/latex]
Let [latex]u=1+2{x}^{3},[/latex] so [latex]du=6{x}^{2}dx.[/latex]
Since the original function includes one factor of [latex]x^2[/latex] and [latex]du=6{x}^{2}dx,[/latex] multiply both sides of the du equation by [latex]\frac{1}{6}.[/latex]
Watch the following video to see the worked solution to this example.
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Let [latex]u=f\left(x\right)[/latex] and [latex]v=g\left(x\right)[/latex] be functions with continuous derivatives on [latex]\left[a,b\right][/latex]. Then
Find the area of the region bounded above by the graph of [latex]y={\tan}^{-1}x[/latex] and below by the [latex]x[/latex] -axis over the interval [latex]\left[0,1\right][/latex].
This region is shown in Figure 1. To find the area, we must evaluate [latex]\underset{0}{\overset{1}{\displaystyle\int }}{\tan}^{-1}xdx[/latex].
Figure 1. To find the area of the shaded region, we have to use integration by parts.
For this integral, let’s choose [latex]u={\tan}^{-1}x[/latex] and [latex]dv=dx[/latex], thereby making [latex]du=\frac{1}{{x}^{2}+1}dx[/latex] and [latex]v=x[/latex]. After applying the integration-by-parts formula we obtain
At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since [latex]\frac{\pi }{4}-\frac{1}{2}\text{ln}2\approx 0.4388[/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.
Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\left(x\right)={e}^{\text{-}x}[/latex], the x-axis, the y-axis, and the line [latex]x=1[/latex] about the y-axis.
The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).
Figure 2. We can use the shell method to find a volume of revolution.
To find the volume using shells, we must evaluate [latex]2\pi {\displaystyle\int }_{0}^{1}x{e}^{\text{-}x}dx[/latex]. To do this, let [latex]u=x[/latex] and [latex]dv={e}^{\text{-}x}[/latex]. These choices lead to [latex]du=dx[/latex] and [latex]v={\displaystyle\int }^{\text{ }}{e}^{\text{-}x}=\text{-}{e}^{\text{-}x}[/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain
Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[/latex] and height of [latex]\frac{1}{e}[/latex] added to the volume of a cone of base radius [latex]1[/latex] and height of [latex]1-\frac{1}{3}[/latex]. Consequently, the solid should have a volume a bit less than
