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Find $f(x)$.$$f^{\prime}(x)=2 e^{x}-6 \sqrt{x}+2, f(0)=4$$

$$2 e^{x}-4 x^{3 / 2}+2 x+2$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 2

Applications of Antidifferentiation

Integrals

Harvey Mudd College

Baylor University

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

00:51

Let $f(x)=x^{2}+4 x-2 .$ F…

00:47

00:37

Find $f(x)$ at the indicat…

01:42

Find $f^{\prime}(x)$ for t…

01:28

Find $f^{\prime}(x)$ if $f…

00:52

Find $f$ . $f^{\prime …

01:39

Find $f^{\prime}(x)$.$…

01:18

Find $f$$$f^{\prime}(x…

we want to find ffx given F prime of x equals two. E V x minus six square root X plus two. Where F of zero is equal to four. This question is challenging us to solve a differential equation for F. We're giving specifically initial value problems such that we want to solve our F explicitly. Rather than using a standard technique to solve this differential equation, we'll take advantage of the fact that F prime of X is simply a function of X means that we can take f crimes anti derivative and solve for the initial conditions to find F. So taking the anti derivative, we have fx equals two. E V x minus four extra three half plus two X plus C. Where C is a constant of integration. Next we use our initial conditions to solve for C. Four equals zero equals C. This is because all of our X terms for zero simply vanish. So C is simply four. This means that we have final solution given at the bottom F of x equals two. E V x minus four extra three halfs plus two X plus four.

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