1.7 Integration using substitution
When we know an integral
∫ e x d x = e x + c \int e^x \d{x} = e^x + c ∫ e x d x = e x + c we can evaluate related integrals like:
∫ e − x d x = − e − x + c \int e^{-x} \d{x} = -e^{-x} + c ∫ e − x d x = − e − x + c by making a substitution . Specifically, we define a new variable u ,
calculate its differential, and then replace both. For example, in this case,
let u = − x u = -x u = − x d u = − d x \d{u} = -\d{x} d u = − d x x = − u x = -u x = − u d x = − d u \d{x} = -\d{u} d x = − d u
∫ e − x d x = − ∫ e u d u = − e u + c = − e − x + c \int e^{-x} \d{x} = -\int e^u \d{u} = -e^u + c = -e^{-x} + c ∫ e − x d x = − ∫ e u d u = − e u + c = − e − x + c Let’s do a more complicated example:
∫ x e x 2 d x \int x e^{x^2} \d{x} ∫ x e x 2 d x Let u = x 2 u=x^2 u = x 2 d u = 2 x d x \d{u} = 2x \d{x} d u = 2 x d x
∫ e x 2 x d x = ∫ e u ⋅ 1 2 d u = 1 2 e u + c \begin{align}
\int e^{x^2} x\d{x} = \int e^u \cdot \frac{1}{2}\d{u} = \frac{1}{2} e^u + c
\end{align} ∫ e x 2 x d x = ∫ e u ⋅ 2 1 d u = 2 1 e u + c Plugging u back in gives the final answer:
∫ x e x 2 d x = 1 2 e x 2 + c \int xe^{x^2} \d{x} = \frac{1}{2}e^{x^2} + c ∫ x e x 2 d x = 2 1 e x 2 + c