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Suppose that $ f(1) = 2 $, $ f(4) = 7 $, $ f^\prime(1) = 5 $, $ f^\prime(4) = 3 $ and $ f^{\prime\prime} $ is continuous. Find the value of $ \displaystyle \int_1^4 x f^{\prime\prime} (x)\ dx $.

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02:40

Wen Zheng

06:10

Willis James

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Juste I.

June 21, 2021

Suppose that f(1) = 2, f(4) = 8, f '(1) = 7, f '(4) = 4, and f '' is continuous. Find the value of 4 xf ''(x) dx 1 .

Missouri State University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Suppose that $ f(1) = 2 $,…

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Suppose that $f(1)=2, f(4)…

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$\begin{array}{l}{\text { …

How to solve this question…

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Suppose that f(1) = 3,…

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Suppose $f^{\prime \prime}…

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If $f(1)=12, f^{\prime}$ i…

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Suppose $f$ is continuous,…

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If $ f(1) = 12 $, $ f'…

to evaluate the definite integral from 1 to 4 of x times F double prime of x dx. Given the information here, we would apply integration by parts and here we let u equal to X. And devi would be F double prime of X dx. And so the differential of you would be equal to dx And we would be the integral of deVI which is F prime of X. and so the integral from 1 to 4 of X times F double prime of x dx. This is just U v minus the integral of VD you. This would be extends F prime of x minus the integral of F prime of X. The X The first term evaluated from 1 to 4 And we have this integral from 1-4, integrating further, we would have X times F prime of X Evaluated from 1 to 4 -4 f of X, Evaluated from 1- four. Now we can combine these two and you would get x times f prime of x minus f of X, Evaluated from 1- four. When access for we have four times Ephraim A four -F of four and then when exist one, we would have one times F of one or f rank of one -F of one. This is equal to four times F. Prima for which is equal to three minus Fo four, which is seven and then minus. We have one times F prime of one, which is five -F of one, which is equal to two. Simplifying this, you would get 12 -7 -3 which is equal to two and so this is the value of the definite integral.

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