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Fundamental Theorem of Calculus

There are two parts of the Fundamental Theorem of Calculus:

Part One

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

Part Two

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

Integral with Functions as Bounds

One Bound

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

Two Bounds

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

No Bounds

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

Example

MathJax TeX Test Page $$\text{Calculate } \frac{d}{dx} \int_{x}^{x^3}{sin(t)}\, \mathrm{d}t$$ $$\text{From the second definition, it equals } 3t^2\sin{t^3} - \sin{t}$$
David Witten