Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.
integration by parts for definite integrals
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
