How do you use the Fundamental Theorem of Calculus to find the derivative of int xsin(πx)  from 1 to x^2?

Dec 9, 2016

$\frac{d}{\mathrm{dx}} I = 2 {x}^{3} \cdot \sin \left(\pi \cdot {x}^{2}\right)$

Explanation:

Strictly speaking you don't, because you can't have the variable you're integrating show up in the limits of integration. Assuming you meant something like

$I = {\int}_{1}^{{x}^{2}} t \sin \left(\pi t\right) \mathrm{dt}$

We see that the fundamental theorem of calculus states that

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$

Where $f \left(x\right) = \frac{d}{\mathrm{dx}} F \left(x\right)$ so using the Fundamental Theorem of Calculus we have

$I = F \left({x}^{2}\right) - F \left(1\right)$

Now, we just differentiate both sides

$\frac{d}{\mathrm{dx}} I = \frac{d}{\mathrm{dx}} F \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} F \left(1\right)$

$F \left(1\right)$ is a constant so its derivative is $0$

$\frac{d}{\mathrm{dx}} I = \frac{d}{\mathrm{dx}} F \left({x}^{2}\right)$

Using the chain rule:

$u = {x}^{2}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

$\frac{d}{\mathrm{dx}} I = \frac{d}{\mathrm{du}} F \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} = 2 x \cdot \frac{d}{\mathrm{du}} F \left(u\right)$

However we know that $F \left(u\right)$ is a function such that its derivative is $u \cdot \sin \left(\pi \cdot u\right)$

$\frac{d}{\mathrm{dx}} I = 2 x \cdot u \cdot \sin \left(\pi \cdot u\right)$

But since $u = {x}^{2}$

$\frac{d}{\mathrm{dx}} I = 2 x \cdot {x}^{2} \cdot \sin \left(\pi \cdot {x}^{2}\right) = 2 {x}^{3} \cdot \sin \left(\pi \cdot {x}^{2}\right)$