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How do you use the Fundamental Theorem of Calculus to find the derivative of #int 6*(sin(t))^2 dt# from 1 to e^x?

Answer 1

Cameron Alexander

#= 6 e^x (sin(e^x))^2 #

we have #d/dx \ int_1^(e^x) \ 6*(sin(t))^2 \ dt#

STEP 1: FTC pt2 states that

#d/(du) int_a^u \ f(t) \ dt = f(u)#

STEP 2: from the chain rule #d/dx = d/(du) * (du)/dx#

So #d/(dx) int_a^(u(x)) \ f(t) \ dt = f(u) * (du)/dx#

here that means that

#d/dx \ int_1^(e^x) \ 6*(sin(t))^2 \ dt = 6*(sin(e^x))^2 d/dx e^x#

#= 6 e^x (sin(e^x))^2 #

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Answer 2

Emilia Aguilar

To find the derivative of ( \int_{1}^{e^x} 6\sin^2(t) , dt ) using the Fundamental Theorem of Calculus, we can first express it as an antiderivative. Then, we differentiate the result with respect to ( x ).

Let ( F(x) = \int_{1}^{e^x} 6\sin^2(t) , dt ).

By the Fundamental Theorem of Calculus, ( F'(x) = 6\sin^2(e^x) \cdot (e^x)' ).

Using the chain rule, ( (e^x)' = e^x ).

Thus, ( F'(x) = 6\sin^2(e^x) \cdot e^x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.