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Which of the following integrals are improper? Why?
(a) $ \displaystyle \int_0^{\frac{\pi}{4}} \tan x\ dx $(b) $ \displaystyle \int_0^\pi \tan x\ dx $(c) $ \displaystyle \int_{-1}^1 \frac{dx}{x^2 - x - 2} $(d) $ \displaystyle \int_0^\infty e^{-x^3}\ dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 8
Improper Integrals
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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So for a look at the graph of tante, first interval from 0 to pi over 4 is finite and there's a function with tan and x has no infinite discontinuity of the integral from there to pi over 4 point. So it is not improper for be also located. The graph of tenants we will find 10 into pi over 2 is infinite, so canon x has an infinite discontinuity on this interval from there to pi, so it is improper integral for c, where x goes to negative 1. The denominator of this function x, squared minus x, minus 2, goes to 0 point, so this function goes to infinite, while x goes to negative yit, so it is improper, for the base. Interval is infinite, so it is improper inte.
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