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Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

$ \displaystyle \int_1^\infty \frac{\ln x}{x^2}\ dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 8

Improper Integrals

Integration Techniques

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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 why the h integral is converted or divergent and the evaluated those that are converted for this improper, integral by definition. This is equal to the limit, a goes to infinity and integral from a 1 to a of the function of x over x. Squared dx we compute is definite integral first, so for this definite integral this is equal to so here we use the method of integration by parts. This is equal to 1 from 1 to a the function is n x and d negative 1 over x. So this is equal to is equal to x times negative 1 over x from 1 to it, minus integral of 1 to a of the function negative 1 over x times 1 over x x. Here we use the method integration by parts, so this is equal to n x times m 1 over x from 1 to a minus. Here, we kind of pat integral from 1 to a 1 over x, squared dx having a and want these 2 functions. This is equal to n a times negative 1 over a minus 0 plus second part is negative 1 over x from 1 to a this is equal to m a times negative 1 over as negative 1 over a minus negative 1. When a goes to infinity. This function goes to 0 and this function also goes to 0. So the answer is this: improper, integral is converted to hyde, is 1.

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