Question

How do you use the first part of the Fundamental Theorem of Calculus to find the derivative of `y = int 3(sin(t))^4 dt` from `e^x` to 1?

Answer 1

The derivative is `-3e^x(sin(e^x))^4`.

Explanation:

`y = int_(e^x)^(1)3(sin(t))^4dt`

Let's let `f(t) = 3(sin(t))^4`.

Remember that by the fundamental theorem of calculus, a definite integral of a function is calculated using its antiderivative.

`int_a^b f(t) dt = F(b) - F(a)` where `F` is an antiderivative of `f`.

Similarly:

`int_(e^x)^1 f(t)dt = F(1) - F(e^x)` where `F` is an antiderivative of `f`.

`y = F(1) - F(e^x)`

Because the question asks for it, we take the derivative.

`color(red)(d/dx)y = color(red)(d/dx)(F(1)) - color(red)(d/dx)(F(e^x))`

Don't be scared. `F(1)` is just a constant.

`dy/dx = 0 - F'(e^x)*d/dx(e^x)`

But by definition of an antiderivative, `F'(x) = f(x)`.

`dy/dx = -e^xf(e^x)`

`dy/dx = -e^x(3(sin(e^x))^4)`

`dy/dx = -3e^x(sin(e^x))^4`