is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Integrals with Functions as Bounds

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. 


MathJax TeX Test Page $$\text{Calculate } \frac{d}{dx} \int_{x}^{x^3}{sin(t)}\, \mathrm{d}t$$ 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: $$\dfrac{d}{dx}\left(F(x^2) - F(\sqrt{x})\right)$$ $$2xF'(x^2) - \dfrac{1}{2\sqrt{x}}*F'(\sqrt{x})$$ $$2x*\sqrt{x^2}\cos(x^2) - \dfrac{1}{2\sqrt{x}}x^{1/4}\cos(\sqrt{x})$$ $$\boxed{2x^2\cos(x^2) - \frac{1}{2}x^{-1/4}\cos(\sqrt{x})}$$
David Witten

Taylor Polynomials

Testing for Convergence and Divergence