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Evaluate the integral.

$ \displaystyle \int_1^2 w^2 \ln w dw $

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$\int_{1}^{2} w^{2} \ln w d w=\frac{8 \ln 2}{3}-\frac{7}{9}$

03:04

Wen Zheng

05:48

Willis James

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Campbell University

Harvey Mudd College

Baylor 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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So here we are given an integral. But first let's ride it with a definite integral. But first let's write an indefinite integral for So right away this is a form of integration by parts. Since we don't know the integral of natural UNCW we have to take this as you will you take uh D V as equivalent to w squared t W. So U equals natural look of W Do. You would be equivalent to one over W. The W And fee would be equivalent to w cubed over three. And now we have sufficient information. Remember integration by parts you? D V equals U V minus V. D. U. So this would be equivalent to W cube Over three. Natural log W. So the minus V. D. U. So this would be equivalent to minus the integral of W. Mhm. G W. So this is equivalent to WQ over three. Natural log of W minus w squared over two. And let's make this easier by factoring out a w squirt. So this would be equivalent to W Over three. Natural log of W minus one house. And now let's put this in terms of a definite role which we were initially interested in. So lower limit and upper limit of integration. So this would be equivalent to four Times 2/3 Natural log of two minus one half. Then this is attracted by that the squared would just be one in this case. So it b minus one third. Natural lack of one is equivalent to zero. So this this crosses out And this would be -1 half. So you see this is 8/3 natural log of two. This would be minus two plus one half, Which is -3 house and this gives our solution to the definite integral.

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