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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. F(x) =...

Question:

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

{eq}F(x) = \int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t {/eq}

Definite Integrals:

The fundamental formula that is used to find the derivative of the integral expression of the form: {eq}\int_{k(x)}^{a} f(t) dt {/eq} will be given as : {eq}\frac{d\left ( \int_{k(x)}^{a} f(t) dt \right )}{dx}=-\frac{d(k(x))}{dx} f(k(x)) \\ {/eq} The derivative of the integral is used with the theory of limits too, but for the derivative of the integral expression, we use the formula given above.

Answer and Explanation:


In this question we are having the integral expression with lower limits as a variable term.

So we have an integral expression:

{eq}F(x) = \int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t {/eq}

The derivative will be found as follows:

{eq}\frac{d(F(x))}{dx}= \frac{d\left (\int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t \right )}{dx}\\ {/eq}

There is fundamental theorem that is used here as per the formula:

{eq}\frac{d\left ( \int_{k(x)}^{a} f(t) dt \right )}{dx}=-\frac{d(k(x))}{dx} f(k(x)) \\ {/eq}

So the given expression will be now;

{eq}\frac{d(F(x))}{dx}= \frac{d\left (\int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t \right )}{dx}\\\\ =- \frac{d(x)}{dx} \sqrt{1 + \sec(8x)} \\ =-\sqrt{1 + \sec(8x)}\\ \Rightarrow F'(x)= -\sqrt{1 + \sec(8x)}\\ {/eq}

So this is the required derivative.


Learn more about this topic:

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Definite Integrals: Definition

from Math 104: Calculus

Chapter 12 / Lesson 6
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