Math 22 Integration by Parts and Present Value

Integration by Parts

 Let $u$ and $v$ be differentiable functions of $x$.
 
 $\int udv=uv-\int vdu$

Exercises Use integration by parts to evaluation:

1) $\int \ln xdx$

Solution:
Let $u=\ln x$, $>du={\frac {1}{x}}dx$
and $dv=dx$ and $v=x$
Then, by integration by parts: $\int \ln xdx=x\ln x-\int x{\frac {1}{x}}dx=x\ln x-\int dx=x\ln x-x+C$

2) $\int xe^{3x}dx$

Solution:
Let $u=x$, $du=dx$
and $dv=e^{3x}dx$ and $v={\frac {1}{3}}e^{3x}$
Then, by integration by parts: $\int xe^{3x}dx=x{\frac {1}{3}}e^{3x}-\int {\frac {1}{3}}e^{3x}dx=x{\frac {1}{3}}e^{3x}-{\frac {1}{9}}e^{3x}+C$

3) $\int x^{2}e^{-x}dx$

Solution:
Let $u=x^{2}$, $du=2xdx$
and $dv=e^{-x}dx$ and $v=-e^{-x}$
Then, by integration by parts: $\int x^{2}e^{-x}dx=x^{2}(-e^{-x})-\int -e^{-x}2xdx=-x^{2}e^{-x}+\int 2xe^{-x}dx$
Now, we apply integration by parts the second time for $\int 2xe^{-x}dx$
Let $u=2x$, $du=2dx$
and $dv=e^{-x}dx$ and $v=-e^{-x}$
So $\int 2xe^{-x}dx=2x(-e^{-x})-\int -e^{-x}2dx=-2xe^{-x}-e^{-x}+C$
Therefore, $\int x^{2}e^{-x}dx=-x^{2}e^{-x}-2xe^{-x}-e^{-x}+C$

Note

1. Tabular integration technique (look it up) is convenient in some cases.

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