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Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves about the given axis.

$ y = \cos (\frac{\pi x}{2}) $ , $ y = 0 $ , $ 0 \le x \le 1 $ ; about the y-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

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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Use the method of cylindri…

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02:37

The problem is used the method of its cylindrical shouts to find the volume generated by rotating the region bond the proves at the given axis y is equal to sine pi x. Over 2 and y is equal to 0 x is between 0 and 1 batiste. First, that we can go the graph is as follows: this is from 0 to 1 and this the graph of y is equal to sine pi x over 2 point. So the region is this part and that we we need to rotate this region about y axis and find the volume so use. Selindrical shells method we can see volume at v is equal to 2. Pi is integral from 0 to 1 times x, times f of x, but this is cosine pi x over 2 dx. Now for this problem we can use integration by parts, so the formula is integral from a to b. You prime d x is equal to. U times from a to b minus integral from a to b, o? U s now for our problem, we can right? U is equal to x and prime is equal to cosine pi x. Over 2 point, then: u, prime, is equal to 1 and v is equal to 2 over pi times sine pi x over 2 point now this is equal to 2 pi times by this formula times. This is r 2 over pi times x times sine pi x. Over 2 from 0 to 1 and minus integral of prime times v say this is 2 over pi sine pi x over 2 from 0 to 1 d. No, this is equal to this is equal to 2 pi times pi over 2 plugging 10 to x. I this is equal to 1 and then minus 2 over pi times the integral of sine pi x over 2. This is equal to negative cosine, which is cosine plus 2 over pi cosine pi x over 2 from 0 to 1 point. This is equal to 2 pi times pi 2 over pi, plus all over pi square times. Cosine pi over 2 is 0 and cosine. 0 is 1, since the answer is 4 minus 8 over pi. This is a.

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