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Find $f(x)$.$$f^{\prime \prime}(x)=\frac{2 x^{2}-1}{x^{4}}, f^{\prime}(1)=3, f(2)=1$$

$$-2 \ln |x|-\frac{1}{6 x^{2}}+\frac{14}{3} x-\frac{199}{24}+2 \ln 2$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 2

Applications of Antidifferentiation

Integrals

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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.

01:53

Find $f$.$$f^{\prime \…

01:02

Find $f(x)$.$$f^{\prim…

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01:27

Find $f^{\prime}(x)$ at $x…

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Find $f^{\prime}(x)$.$…

Find $f^{\prime}(x)$$$…

05:10

First find $f^{\prime}$ an…

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Find $$f^{\prime}(a)$$…

01:52

we want to find a Fedex given F double prime of X is equal to two over X squared minus one over X to the fourth. This is a second order differential equation with initial value problem or initial conditions that time of one equals three. F two equals one. Since F double prime of X is simply a function of X. We can take anti deliver successively plugging our initial conditions to solve for fx. So first our first anti derivative gives F prime X equals negative two over X plus 1/3 X cubed plus the cost of integration. See plugging in F prime one equals three, gives us three equals negative two plus one third plus C. Or C equals 40. Number three. Now we take the anti derivative of our first anti derivative to solve for F. This gives on the first line negative to L N x minus 1/6 X squared plus 14/3 X plus D. The second class of integration, plugging in F two equals one gives D equals negative 199 over 24 plus two. Ln two. Thus we have our final solution, F equals negative two. L n x minus 1/6 X squared plus 14/3 x minus 1 99. Number 24 plus two L n two.

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