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

Answer:

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#