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Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
$ \displaystyle \int_1^\infty \frac{1}{(2x + 1)^3}\ dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 8
Improper Integrals
Integration Techniques
Oregon State University
Harvey Mudd College
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.
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Determine whether each int…
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The problem is determine whether this integral has converted or divergent. This is an improper integral the first we can use so that, u is equal to 2 x plus 1 to convert this function, to a function that easier to integral a is equal to 2 x, plus, 1 and d. U is equal to 2 times dexter, this integral is equal to integral 1 over. U 23 is the power, and dx is equal to 1 half du since du is equal to 2 times x, and this is from 3 in x, equal to 1. U is equal to 3 and when x goes to infinity, you also goes to infinity and then, by definition, this is equal to the limit. He goes to infinity and integral 1 over: u to 3 power du from 3 to the distance equal to the limit. He goes to infinity and anti derivative of 1 over. U to 3 power is equal to 1 over negative 3 plus 1 times. U to negative 3 cos la from 3 to t, then plotting t and 3. Here this is equal to limit. She goes to infinity won over negative 2 times negative 2 minus 3 to negative 2 and when t goes to infinity, he to negative choose power goes to 0 point, so the answer is equal to negative 1 half times negative 32 negative 2 powers. So this is 1 half times 1 over 9 point. This is 1 over 18 point, so this integral is converted and the value is 118.
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