Respuesta :
Answer:
The Volume is 5.018 cubic units
Explanation:
Volume Of A Solid Of Revolution
Let f(x) be a continuous function defined in an interval [a,b], if we take the area enclosed by f(x) between x=a, x=b and revolve it around the x-axis, we get a solid whose volume can be computed as
[tex]\displaystyle V=\pi \int_a^bf^2(x)dx[/tex]
It's called the disk method. There are other available methods to compute the volume.
We have
[tex]f(x)=xe^x[/tex]
And the boundaries defined as x=1, y=0 and revolved around the x-axis. The left endpoint of the integral is easily identified as x=0, because it defines the beginning of the region to revolve. So we need to compute
[tex]\displaystyle V=\pi \int_0^1(xe^x)^2dx=\pi \int_0^1x^2e^{2x}dx[/tex]
We need to first determine the antiderivative
[tex]\displaystyle I=\int x^2e^{2x}dx[/tex]
Let's integrate by parts using the formula
[tex]\displaystyle \int u.dv=u.v-\int v.du[/tex]
We pick [tex]u=x^2,\ dv=e^{2x}dx[/tex]
Then [tex]du=2xdx,\ v=\frac{e^{2x}}{2}[/tex]
Applying by parts:
[tex]\displaystyle I=x^2\frac{e^{2x}}{2}-\int 2x\frac{e^{2x}}{2}dx[/tex]
[tex]\displaystyle I=\frac{x^2e^{2x}}{2}-\int xe^{2x}dx[/tex]
Now we solve
[tex]\displaystyle I_1=\int xe^{2x}dx[/tex]
Making [tex]u=x,\ dv=e^{2x}dx[/tex]
[tex]\displaystyle du=dx,\ v=\frac{e^{2x}}{2}[/tex]
Applying by parts again:
[tex]\displaystyle I_1=x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}dx[/tex]
[tex]\displaystyle I_1=\frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx[/tex]
The last integral is directly computed
[tex]\displaystyle \int e^{2x}dx=\frac{e^{2x}}{2}[/tex]
Replacing every integral computed above
[tex]\displaystyle I=\frac{x^2e^{2x}}{2}-\left(\frac{xe^{2x}}{2}-\frac{1}{2}\frac{e^{2x}}{2}\right)[/tex]
Simplifying
[tex]\displaystyle I=\dfrac{\left(2x^2-2x+1\right)\mathrm{e}^{2x}}{4}[/tex]
Now we compute the definite integral as the volume
[tex]V=\pi \left[\dfrac{\left(2(1)^2-2(1)+1\right)\mathrm{e}^{2(1)}-\left(2(0)^2-2(0)+1\right)\mathrm{e}^{2(0)}}{4}\right][/tex]
Finally
[tex]V=\pi \dfrac{\mathrm{e}^2-1}{4}=5.018[/tex]
The Volume is 5.018 cubic units