All My Bloody Notes Bloody Maths Calculus Integration Rules

Integration Rules

You can't get away with not knowing integration by-parts

Integration by-parts

udv=uvvdu\int u dv = uv - \int v du

The key is to find these expressions:

  1. dudu: So you need to find a uu that can be easily differentiated
  2. vv: So you need to find a dvdv that can be easily integrated


E(X)=0wxf(x)dxE(X) = \int_{0}^{w} x f(x) dx

Assign uu as xx and dvdv as f(x)f(x), then we have:

  1. dudx=1    du=dx\frac{du}{dx} = 1 \implies du = dx
  2. dv=v    f(x)dx=F(x)\int dv = v \implies \int f(x) dx = F(x)

Then we have: E(X)=0wxf(x)dx=xF(x)0wF(x)dxE(X) = \int_{0}^{w} x f(x) dx = xF(x) - \int_{0}^{w} F(x) dx

Even quicker:

Notice that you can rewrite 0wxf(x)dx=0wxdF(x)\int_{0}^{w} x f(x) dx = \int_{0}^{w} x dF(x)

Why? Because dF(x)dx=f(x)\frac{dF(x)}{dx} = f(x) or dF(x)=f(x)dxdF(x) = f(x) dx

Then you can apply integration by-parts directly with substituting u=xu = x and v=F(x)v = F(x) in the original expression:

0wxdF(x)=xF(x)0wF(x)dx\int_{0}^{w} x dF(x) = x F(x) - \int_{0}^{w} F(x) dx

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