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Evaluate the integral.

$ \displaystyle \int_0^{\frac{1}{2}} x \cos \pi x dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Campbell University

Oregon State University

Baylor University

University of Nottingham

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.

04:51

Evaluate the integral.…

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Evaluate the integrals.

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The problem is violated as integral integral from there to a half x times, cosine pi x, dx. For this problem we can use the method of integration by parts integral from a to b. U v, prime dx! This is equal to u v from a to b minus the integral of the prime x from a to b now for our problem problem, we can let? U is equal to x and prime is equal to cosine pi x, then prime is equal to 1 and v is equal to 1. Over pi times sine pi x now use integration by parts we have. This is equal to x, times 1 over pi sine pi times x from 0 to 1, half minus the integral of prime times v. So this is 1 over pi sine pi x, dx from 0 to 1 half the first term. This term is equal to this. Is equal to this term plugging a half? This is 1 over 2 pi and minus 0 and then minus the integral of this. 1 is 1 over pi squared times negative cosine pi x from 0 to 1 half pint. This is equal to 1 over 2 pi plus 1 over pi square cosine pi over 2 minus cosine. This is equal to negative 1. So is the answer is 1 over 2 pi minus 1 over pi square.

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