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?

1 Answer
Jul 23, 2018

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