Question

How do you find the derivative of `y= int 6*(sin(t))^2 dt` from a=e^x to b=1?

Answer 1

Use second fundamental theorem of calculus to find the answer. The working is given below.

Explanation:

`y=int_(e^x)^1 6*(sin^2(t))dt`
`y=-int_1^(e^x) 6*(sin^2(t))dt`

`dy/dx = -dy/dx(int_1^(e^x) 6*(sin^2(t))dt)`

Using second fundamental theorem of calculus we replace `t` by `e^x` and multiply the result by derivative of `e^x`

`dy/dx = -(6sin^2(e^x))(e^x)`

`dy/dx = -6e^xsin^2(e^x)` Answer