26 lines
1.1 KiB
Markdown
26 lines
1.1 KiB
Markdown
The integration by parts formula is:
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$$ \int udv = uv - \int vdu $$
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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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$$ \int udv = uv - \int v du $$ |