The Chain Rule

FTC 1

For some continuous function $f(t)$:

$$F(x)={\int }_a^{x}f(t)dt⟹F^{\prime }(x)=\frac{d}{dx}{\int }_a^{x}f(t)dt=f(x)$$

We can substitute the bounds of integration:

$$\frac{d}{dx}{\int }_a^{\tan (x)}{t}^3+4tdt$$

• Let $u=\tan (x)$
  • $\frac{du}{dx}={\sec }^2(x)$

$$\frac{d}{\cancel{du}}\cdot \frac{\cancel{du}}{dx}=\frac{d}{dx}⟹\frac{d}{du}\cdot \frac{du}{dx}{\int }_a^u{t}^3+4tdt$$ $${\sec }^2(x)\frac{d}{du}{\int }_a^u{t}^3+4tdt]={\sec }^2(x)[u^3+4u]$$

• Substitute back $u=\tan (x)$

$${\sec }^2(x)[{\tan }^3(x)+4\tan (x)]$$

We can generalize this as:

$$\frac{d}{dx}{\int }_a^{g(x)}f(t)dt=f(g(x))\cdot g^{\prime }(x)$$

This is the chain rule.

FTC 2

$${\int }_b^{a}f(x)dx=F(b)-F(a)$$

Where $F(x)$ is the antiderivative of $f(x)$. Note that the function must be continuous over the bounds of integration. Otherwise, it is an improper integral.

For instance, consider the function $f(x)=\frac{1}{x^2}$:

$$\int f(x)dx=-\frac{1}{x}+c$$

We can attempt to integrate over $[-2,2]$ using FTC 2:

$${\int }_{-2}^{2}f(x)dx=-\frac{1}{2}-\frac{1}{2}=-1$$

However, this integral is not equal to $-1$, as $f(x)$ is discontinuous at $x=0$, which is contained within the bounds of integration. Note that the function continues to infinity, implying an undefined area.

Therefore,

$${\int }_{-2}^{2}f(x)dx\ne F(-2)-F(2)$$