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8.4 Improper Integrals AP Calculus BC

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8.4 Improper Integrals One of the great characteristics of mathematics is that mathematicians are constantly finding ways to get around the rules, or to bend the rules, or just plain ignore them. Improper integrals is a technique to use when an interval is not finite, and when an integrand is not continuous.

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Until now we have been finding integrals of continuous functions over closed intervals. Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals.

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First, here are the definitions of Improper Integrals with Infinite Solutions: If the Limit is finite, then the Improper Integral converges. If the Limit fails, then it diverges If both Improper Integrals converge, then so does

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Express the improper integral as a limit of definite integrals and evaluate the integral. First, rewrite and split up the integral. The antiderivative of = 0 as b → ∞

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Express the improper integral as a limit of definite integrals and evaluate the integral. Now, by analyzing the integrand, the denominator is always positive for all x. [i.e., (x 2 + 1) 2 is always positive.] So, is positive when x > 0 and negative when x < 0. Therefore, because x < 0 and, = −1 + 1 = 0

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Evaluate the following integral or state that it diverges. Use Integration by Parts = 0

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Improper Integrals with Infinite Discontinuities Integrals of functions that become infinite at a point within the interval of integration are improper integrals. Another type of improper integral arises when the integrand has a vertical asymptote. And that’s also a point of infinite discontinuity either at a limit of integration or at some other point between the limits of integration. Finite Limit Converges Limit Fails Diverges Both Limits Finite Converges

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Example 1: The function is undefined at x = 1. Since x = 1 is an asymptote, the function has no maximum. Can we find the area under an infinitely high curve? We could define this integral as: (left hand limit) We must approach the limit from inside the interval.

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Rationalize the numerator.

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This integral converges because it approaches a solution.

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Example 2: This integral diverges. (right hand limit) We approach the limit from inside the interval.

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Example 3: The function approaches when.

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Example 4: What happens here? If then gets bigger and bigger as, therefore the integral diverges. If then b has a negative exponent and, therefore the integral converges. (P is a constant.) Day 1

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Theorem 6Comparison Test

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Converges

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Does converge? Compare: to for positive values of x. For

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Since is always below, we say that it is “bounded above” by. Since converges to a finite number, must also converge!

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Example 7: The maximum value of so: on Since converges, converges.

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Example 7: for positive values of x, so: Since diverges, diverges. on

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If functions grow at the same rate, then either they both converge or both diverge. Does converge? As the “1” in the denominator becomes insignificant, so we compare to. Since converges, converges.

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Of course

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Day 2

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