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Find $f(x)$.$$f^{\prime}(x)=2+5 / x, f(1)=3$$

$$2 x+5 \ln |x|+1$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 2

Applications of Antidifferentiation

Integrals

Missouri State University

Oregon State University

Harvey Mudd College

Boston College

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.

01:06

Find $f(x)$.$$f^{\prim…

01:02

01:42

Find $f^{-1}.$$$f(x)=\…

0:00

Find $f^{\prime}(a)$.$…

01:23

Find $f^{\prime}(x)$.$…

02:07

Find $f^{\prime \prime}(x)…

01:00

find $f^{\prime \prime}(2)…

01:01

Find $f^{\prime}(x)$$$…

01:17

Find $f$$$f^{\prime}(x…

00:44

Find $f(3)$ if $f(x)=-4 x^…

we want to find F of X given F prime of X is equal to two plus five over X. For the initial conditions F of one is equal to three. This question is challenging us to solve a differential equation for the function F of X. Specifically, it's asking us to solve the initial value problem given dissolve this iVP. Rather than using a standard method for solving differential equations. Since F prime isn't fully in terms of X, that is, there's no why on the right to the equation, we can simply take the anti derivative F prime. And these are initial conditions to solve. So first let's take the anti derivative. This gives us as equals two X plus five. Natural algorithm of X plus a constant of integration. See next result for C. Using F one equals three, so three is equal to two plus five. Ln one plus C l n +10 So C plus two equals three, or C equals one. So we have our final solution fx is equal to two X plus five times the national algorithm of X plus the constant one.

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