notes/education/math/MATH1220 (calc II)/Integration by Parts.md

The integration by parts formula is:

\int udv = uv - \int vdu

Broadly speaking, integration by parts is done by:

1. Pick a part of integral to be u.
2. The rest of the integral will be dv,
3. Compute the derivative of u, du.
4. Compute the antiderivative of dv
5. Substitute those values in to the integration by parts formula.

Deriving the Integration by Parts Formula

\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

1. Integrating both sides, we get:

\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]

2. Through the distributive property of integrals,

= \int f'(x)g(x)dx + \int f(x)g'(x)dx

3. An integral cancels out an antiderivative, therefore:

f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx

4. Moving terms around:

\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx

Now, let u = f(x) and v = g(x), then dv = g'(x)dx and du = f'(x)dx.

Examples

Evaluate the below antiderivative using integration by parts.

\int xe^{2x}dx

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.
   • u = x
   • du = \frac{d}{dx}(x)= 1dx
   • dv = e^{2x}dx
   • v = \frac{1}{2}e^{2x} - The antiderivative of dv.
2. Looking back at the integration by parts formula, we know that: \int udv = uv - \int v du $ \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)-\int (\frac{1}{2}e^{2x}) (1dx) $
3. The remaining integral can be solved with u substitution, but we've already defined u, so we use w as a replacement.
   • w = 2x
   • dw = 2dx
   • \frac{1}{2}dw=dx
     1. Substituting w and dw into the integral:
     int \frac{1}{2}e^w \frac{1}{2}dw $$
     2. This gives an integral that can be computed naively
     int\frac{1}{2}e^{w}\frac{1}{2}dw = \frac{1}{4}\int e^w dw= \frac{1}{4}e^{2x} $$
4. Combining everything together, we get: \int xe^{2x}dx = (\frac{1}{2}e^{2x})(x)- (\frac{1}{4}e^2x) + C$