These problems originally appeared in my former alternate blog at http://www.optimal-learning-systems.org/eps.
Q1. What is the value of the integral \(\int xe^xdx \)?
A1. Let \(u=x\). Then \(du=dx\). Also let \(dv=e^xdx\). Then \( v= \int e^xdx=e^x \).
Now apply the integration by parts formula, \( \int u dv=uv−\int v du \).
We have, \( \int x e^xdx=x e^x−\int e^xdx=x e^x−e^x=e^x[x−1] \).
Therefore, \( \int xe^xdx=e^x[x−1]\).
One may wonder why the constant of integration is not added in the formula for v. The IBP formula becomes in this case \( u(v+C)−∫(v+C)du\) and one can see that the terms containing the constant of integration cancels out.
Q2. Determine the integral \(\int x^2e^x dx\).
A2. Let \(u=x^2\), then \(du=2^xdx\). Also let \(dv=e^xdx\), then \(v=\int e^xdx\). Then \( dv=e^x dx\). Substituting in the IBP formula,
\( x^2e^x−\int e^x(2x)dx=x^2e^x−2∫xe^xdx\). But the last term was the problem we treated before! Hence, \( \int x^2e^xdx=x2e^x−2[e^x(x−1)]\) and therefore
\[\int x^2e^xdx=e^x [x2−2(x−1)]\]
You may add the constant of integration in the final answer.
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Q3. Determine the integral \(\int x^3e^x dx\).
A3. As usual, let \(u=x^3\), then \(du=3x^2dx\). Also let \(dv=e^xdx\), then \(v=e^x\). Substituting in the IBP formula,
\(x^3e^x−\int e^x(3x^2)dx=x^3e^x−3∫x2e^xdx\). But the last term involved an integral we treated before! Therefore \[ \int x^3e^x dx=x^3e^x−3e^x[x^2−2(x−1)]=e^x[x^3−3(x^2−2(x−1))]\]
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Notice that \[ \int x^ne^x dx=x^ne^x−n\int x^{n−1}e^xdx \]. We leave the fun to the student to discover on his/her own a general formula for the answer.
