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

1 Answer
Dec 9, 2016

d/dxI = 2x^3*sin(pi*x^2)

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)tsin(pit)dt

We see that the fundamental theorem of calculus states that

int_a^bf(x)dx = F(b) - F(a)

Where f(x) = d/dxF(x) so using the Fundamental Theorem of Calculus we have

I = F(x^2) - F(1)

Now, we just differentiate both sides

d/dxI = d/dxF(x^2) - d/dxF(1)

F(1) is a constant so its derivative is 0

d/dxI = d/dxF(x^2)

Using the chain rule:

u = x^2 and (du)/dx = 2x

d/dxI = d/(du)F(u)*(du)/dx = 2x*d/(du)F(u)

However we know that F(u) is a function such that its derivative is u*sin(pi*u)

d/dxI = 2x*u*sin(pi*u)

But since u = x^2

d/dxI = 2x*x^2*sin(pi*x^2) = 2x^3*sin(pi*x^2)