42 lines
1.9 KiB
Markdown
42 lines
1.9 KiB
Markdown
The integration by parts formula is:
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$$ \int udv = uv - \int vdu $$
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Broadly speaking, integration by parts is done by:
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1. Pick a part of integral to be $u$.
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2. The rest of the integral will be $dv$,
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3. Compute the derivative of $u$, $du$.
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4. Compute the antiderivative of $dv$
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5. Substitute those values in to the integration by parts formula.
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## Deriving the Integration by Parts Formula
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$$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
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1. Integrating both sides, we get:
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$$\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]$$
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2. Through the distributive property of integrals,
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$$ = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
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3. An integral cancels out an antiderivative, therefore:
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$$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
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4. Moving terms around:
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$$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$
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Now, let $u = f(x)$ and $v = g(x)$, then $dv = g'(x)dx$ and $du = f'(x)dx$.
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# Examples
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> Evaluate the below antiderivative using integration by parts.
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$$\int xe^{2x}dx$$
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1. Define $u$ to be a value you can take the derivative of easily, in this case $u = x$. The rest of the integral will be set to $dv$, in this case, $dv = e^{2x}dx$.
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- $u = x$
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- $du = \frac{d}{dx}(x)= 1dx$
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- $dv = e^{2x}dx$
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- $v = \frac{1}{2}e^{2x}$ - The antiderivative of $dv$.
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2. Looking back at the integration by parts formula, we know that:
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$$ \int udv = uv - \int v du $$
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$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)-\int (\frac{1}{2}e^{2x}) (1dx) $$
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3. The remaining integral can be solved with $u$ substitution, but we've already defined $u$, so we use $w$ as a replacement.
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- $w = 2x$
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- $dw = 2dx$
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- $\frac{1}{2}dw=dx$
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1. Substituting $w$ and $dw$ into the integral:
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$$ \int \frac{1}{2}e^w \frac{1}{2}dw $$
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2. This gives an integral that can be computed naively
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$$ \int\frac{1}{2}e^{w}\frac{1}{2}dw = \frac{1}{4}\int e^w dw= \frac{1}{4}e^{2x} $$
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4. Combining everything together, we get:
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$$ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)- (\frac{1}{4}e^2x) + C$$ |