Is there an inverse chain rule for integration?

Oct 29, 2015

Integration by substitution is the inverse of differentiation using the chain rule.

Explanation:

$\int f \left(g \left(x\right)\right) g ' \left(x\right) \mathrm{dx}$

Let $u = g \left(x\right)$. This make $\mathrm{du} = g ' \left(x\right) \mathrm{dx}$ and the integral becomes

$\int f \left(u\right) \mathrm{du}$

A different way to see this is to do n integration by substitution and then check the answer by differentiation.

Example

$\int 3 {x}^{2} \sqrt{{x}^{3} + 7} \mathrm{dx}$

Let $u = {x}^{3} + 7$. This makes $\mathrm{du} = 3 {x}^{2} \mathrm{dx}$ and the integral becomes:

$\int 3 {x}^{2} \sqrt{{x}^{3} + 7} \mathrm{dx} = \int {\left({\underbrace{{x}^{3} + 7}}_{u}\right)}^{\frac{1}{2}} {\underbrace{3 {x}^{2} \mathrm{dx}}}_{\mathrm{du}} = \int {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{2}{3} {u}^{\frac{3}{2}} + C = \frac{2}{3} {\left({x}^{3} + 7\right)}^{\frac{3}{2}} + C$

Now, if we check our answer by differentiating, we will use the chain rule.

We could also write the integration using $g \left(x\right)$ rather than $u$.

$\int 3 {x}^{2} \sqrt{{x}^{3} + 7} \mathrm{dx} = \int {\left({x}^{3} + 7\right)}^{\frac{1}{2}} 3 {x}^{2} \mathrm{dx}$

With $g \left(x\right) = {x}^{3} + 7$, we have $g ' \left(x\right) = 3 {x}^{2}$, so the integral is

$= \int {\left(g \left(x\right)\right)}^{\frac{1}{2}} g ' \left(x\right) \mathrm{dx} = \frac{2}{3} {\left(g \left(x\right)\right)}^{\frac{3}{2}} + C$

$= \frac{2}{3} {\left({x}^{3} + 7\right)}^{\frac{3}{2}} + C$