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Calculate the volume generated by rotating the region bounded by the curves $ y = \ln x $, $ y = 0 $ and $ x = 2 $ about each axis.(a) The y-axis(b) The x-axis

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Campbell University

Oregon State University

University of Michigan - Ann Arbor

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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Calculate the volume gener…

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Find the volume generated …

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Find the volume obtained b…

The problem is calculated: the volume generated by rotating the region, bounded by the curves y, equal to x y equal to 0 and x, equal to 2 about each axis, is about the y axis is about x axis, so first we can jarrah x and y, and It s there's a graph of the curve y go to x y equals to 0 and x equal to twont, so the region is this part now for a? If we retain this region about y axis, then we can use cylindrical shell method to find the volume is equal to 2 pi, integral from 1 to 2 x times x d x. Then we kind o use method of integration by powers that, u is equal to? U n x and the aprons equal to x, then you prom is equal to 1 over x and the v is equal to 1 half x squared. So this is equal to 2 pi times, u times 3. So this is 1 half x, squared x from 1 to 2 minus the integral from 1 to 2 on your prime times was so. This is 1 half like the x. This is equal to 2 pi times plunging 2 and 1 to this function. This is 2 times in 2, minus 0 and minus 1 fourth x, squared from 1 to 2 point. This is 2 pi times 2 times into minus. This is 3 over 4 foyotthis region about x axis. So we kind of use this kamis is equal to integral from 1 to 2 pi times n x square, then for this integral we can light, can also use integration by parts and that u is equal to x square and is equal to 1. Then your prime is equal to 2 times x, ove x and the prime and is equal to x pot. Now this integral is equal to pi times: u times 3. So this is x, l n x, squared from 1 to 2 minus integral from 1 to 2 priss 2 times x. Now, for this integral we can use integration by pontais equal to u n, x and apron is equal to 1 and prim is equal to 1. Over x and v is equal to x, so this is equal to pi times plug in 2 and 1 to this function. This is 2 into square. Minus 2 times is integration by parts, so this is x and x from 1 to 2 minus integral from 1 to 2. So this is equal to pi 2 times l into square minus 2 times 2 times. N 2 minus this is 1. This is the result.

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