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) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \ln x dx}

Solution:
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\ln x} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle >du=\frac{1}{x}dx}
and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=x}
Then, by integration by parts: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \ln x dx=x\ln x-\int x\frac{1}{x}dx=x\ln x-\int dx=x\ln x -x +C}

2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int xe^{3x}dx}

Solution:
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=dx}
and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=e^{3x}dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, $\int x^{2}e^{-x}dx=-x^{2}e^{-x}-2xe^{-x}-e^{-x}+C$

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