Use Part of the Fundamental Theorem of Calculus to find the derivative derivative of the function y

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asked Dec 17, 2023 8.5k views

1 Answer

Mariuz · Answer 1 · 2023-12-22T21:08:26+0000

Find the derivative of the function:

$y=_{^{}}\int ^(9x)_(-7)(t^2+5)e^(t-2)dt_{}$

The Fundamental Theorem of Calculus I states that if f(x) is continuous over the interval [a,b] and the function F(x) is defined by:

$F(x)=\int ^x_af(t)dt$

Then F'(x) = f(x) over [a,b]

We must find the derivative of y and the upper limit of the integral is not x, but a function of x: g(x) = 9x.

The only difference with the pure definition is that we have to use the chain rule for derivatives as follows:

$y=_{^{}}\int ^(g(x))_(-7)(t^2+5)e^(t-2)dt_{}=((9x)^2+5)e^(9x-2)\cdot g^(\prime)(x)$

Since g'(x) = 9, then:

$y=_{^{}}\int ^(9x)_(-7)(t^2+5)e^(t-2)dt_{}=9((9x)^2+5)e^(9x-2)$

Operating:

$y=_{^{}}\int ^(g(x))_(-7)(t^2+5)e^(t-2)dt_{}=9(81x^2+5)e^(9x-2)$