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Evaluate the integral.

$ \displaystyle \int_1^4 \sqrt{y} \ln y\ dy $

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$\frac{16}{3} \ln 4-\frac{28}{9}$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Integration Techniques

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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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Evaluate the integral.…

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Evaluate the definite inte…

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Evaluate the given definit…

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evaluate the integrals of …

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02:54

Let's use integration by parts for this in a barrel. Let's take you to be natural, Aga boy. Then do you one of her lie D y? And this leaves us with Devi equal squared or why d y and taking the integral here will give us two thirds. Why the three house? So using our formula here This me right on the side, juvie minus and a girl video. So you times V in the wrong order. But there they are And then our end points once before minus in a girl VDO So let me pull out the two thirds And then after cancelling the wise, we have one half. So let's just go ahead and evaluate this term first. So, Ellen, for times two or three and then for the three halfs and then minus zero after you plug in one minus and we've already evaluated this inaugural. So here we should get to over three coming from over here and then another two of three coming from here. Why three house once for so here we have squared off form, which is too then to the third power, which is eight and then times two, so That's sixteen for the first term. And then here. Let's go ahead and simplify this. So we get full over nine and then we got a plug in foreign one. So here, sixteen over three. And then we've already evaluated this to be eight. So take off one and then multiply it by four. So we have twenty eight up there, all divided by nine. How? And that's your final answer.

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