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

The integration by parts formula is:

\int udv = uv - \int vdu

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 $$
     2. This gives an integral that can be computed naively
     int\frac{1}{2}e^{w} $$