Integrals with Functions as Bounds

Fundamental Theorem of Calculus

There are two parts of the Fundamental Theorem of Calculus:

Part One

$$\int_{a}^{b}{f(x)}\, \mathrm{d}x = F(a) - F(b) \text{ where F(x) is the antiderivative of f(x)}$$

Part Two

$$\text{If } F(x) = \int_{a}^{x}\mathrm{f(t)}\, \mathrm{d}t\text{, then } \frac{d}{dx} F(x) = f(x)$$

Integral with Functions as Bounds

One Bound

$$\text{Calculate } \frac{d}{dx} \int_{a}^{h(x)}{f(t)}\, \mathrm{d}t$$ $$\text{Let } G(x) = \int_{a}^{x}{f(t)}\, \mathrm{d}t$$ $$G(h(x)) = \int_{a}^{h(x)}{f(t)}\, \mathrm{d}t$$ $$\frac{d}{dx}G(h(x)) = G'(h(x))h'(x)$$ $$=f(h(x))h'(x)$$

Two Bounds

$$\text{Calculate } \frac{d}{dx} \int_{g(x)}^{h(x)}{f(t)}\, \mathrm{d}t$$ $$\text{Just split it up: } \frac{d}{dx}(-\int_{a}^{g(x)}{f(t)}\, \mathrm{d}t + \int_{a}^{h(x)}{f(t)}\, \mathrm{d}t)$$ $$-f(g(x))g'(x) + f(h(x))h'(x)$$ $$=f(h(x))h'(x) - f(g(x)g'(x)$$

No Bounds

The derivative is 0, because that's just a constant. 

Examples

$$\text{Calculate } \frac{d}{dx} \int_{x}^{x^3}{sin(t)}\, \mathrm{d}t$$ Let F be the antiderivative of sin(t). So, this expression equals $$\dfrac{d}{dx}\left(F(x^3) - F(x)\right)$$ $$3x^2F'(x^3) - F'(x)$$ Recall F is the antiderivative of $\sin(t)$, so $F' = sin(t)$. $$\boxed{3x^2\sin(x^3) - \sin(x)}$$ $$\text{Calculate } \frac{d}{dx} \int_{\sqrt{x}}^{x^2} \sqrt{t}\cos(t)\, \mathrm{d}t$$ Let F be the antiderivative of $\sqrt{t}\cos(t)$. This expression equals: $$\dfrac{d}{dx}\left(F(x^2) - F(\sqrt{x})\right)$$ $$2xF'(x^2) - \dfrac{1}{2\sqrt{x}}*F'(\sqrt{x})$$ $$2x*\sqrt{x^2}\cos(x^2) - \dfrac{1}{2\sqrt{x}}x^{1/4}\cos(\sqrt{x})$$ $$\boxed{2x^2\cos(x^2) - \frac{1}{2}x^{-1/4}\cos(\sqrt{x})}$$