Help with Definite Integral?

Question

Using the Fundamental Theorem of Calculus, how do you find the derivative of sin(t), from x^3 to 1?

THANKS IN ADVANCE!

Answer 1

www.nipissingu.ca/calculus/tutorials/int...

Answer 2

respect to x? and i assume you meant derivative of the INTEGRAL of sin(t) from x^3 to 1. Otherwise, it doesn't really make any sense when you use FTC to compute the limits when you just take a derivative of a function.

if you integrate respect to t, you'll get -cos(t). Compute the limits and you'll get -cos(1) + cos(x³). Now you differentiate respect to x and you'll get -3x² sin(x³)

so

........1

d/dx ∫ sin(t) dt = -3x² sin(x³)

.......x³

Answer 3

d/ dx S sin(t)

Let u= x^3 then du=3x^2 dx

Let y= sin(t) from [1,u] then dy=sin(u) du

So, dy/dx =dy/du*du/dx

3x^2 sin((x^3))