Solved: ?Improper Integrals that Are Both Type 1 and Type 2 The integral
Chapter 7, Problem 68(choose chapter or problem)
Improper Integrals that Are Both Type 1 and Type 2 The integral \(\int_{a}^{\infty} f(x) d x\) is improper because the interval \([a, \infty)\) is infinite. If \(f\) has an infinite discontinuity at \(a\), then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:
\(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x c>a\)
Evaluate the given integral if it is convergent.
\(\int_{2}^{\infty} \frac{1}{x \sqrt{x^{2}-4}} d x\)
Equation Transcription:
Text Transcription:
integral_1^infinity f(x)dx
[a,infinity)
f
a
integral _1 ^infinity f(x) dx = integral_a^infinity f(x) dx = integral_c ^infinity f(x) c>a
integral _1 ^infinity 2+cos x/sqrt x^4+x^2 dx
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