# SOLVED: Comparison Theorem: Suppose f and g are continuous functions with f(x) â‰¥ g(x) â‰¥ 0 for all x in (0,b). Then: (a) If âˆ« f(x)dx converges, then âˆ« g(x)dx also converges. (b) If âˆ« g(x)dx d

Comparison test for convergence and divergence of improper integrals
Comparison test for convergence and divergence of improper integrals

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Comparison Theorem: Suppose f and g are continuous functions with f(x) â‰¥ g(x) â‰¥ 0 for all x in (0,b). Then:
(a) If âˆ« f(x)dx converges, then âˆ« g(x)dx also converges.
(b) If âˆ« g(x)dx diverges, then âˆ« f(x)dx also diverges.
This theorem can be extended to improper integrals of Type arctan * 0.3.
Use the comparison test to determine whether the integral is convergent or divergent: âˆ« dx / âˆš(x^2 + 1)
Hint: Recall that 0 < arctan(x) < Ï€/2 for all x in [0, 6).

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01:41

The following is a Comparison Test for improper integrals. Suppose $f$ and $g$ are continuous and $0 \leq f(x) \leq g(x)$ for $x \geq a$. If $\int_{a}^{\infty} g(x) d x$ converges, then $\int_{a}^{\infty} f(x) d x$ also converges. Use this result to show that the given integral converges.$$\int_{0}^{\infty} e^{-x^{2}} d x$$

01:30

The following is a Comparison Test for improper integrals. Suppose $f$ and $g$ are continuous and $0 \leq f(x) \leq g(x)$ for $x \geq a$. If $\int_{a}^{\infty} g(x) d x$ converges, then $\int_{a}^{\infty} f(x) d x$ also converges. Use this result to show that the given integral converges.$$\int_{0}^{\infty} \frac{1}{x+e^{x}} d x$$

01:40

Find a comparison function for the integrand and determine whether the integral is convergent:

Which of the following is an appropriate comparison function for (1+x^4)/g(x) = g(x)/[g(x)^3(1-x^2)]?

Is the given integral, f(x) = âˆ«(1+x^4)dx, convergent?

OA The given integral is divergent because 0 < g(x) < f(x) for all x^2 and âˆ«g(x)dx is convergent.

The given integral is divergent because 0 < g(x) < f(x) for all x^2 and âˆ«g(x)dx is divergent.

The given integral is convergent because g(x) â‰¥ f(x) for all x^2 and âˆ«g(x)dx is convergent.

The given integral is convergent because g(x) â‰¥ f(x) for all x^2 and âˆ«g(x)dx is divergent.

g(x) =

01:35

The following is a Comparison Test for improper integrals. Suppose $f$ and $g$ are continuous and $0 \leq f(x) \leq g(x)$ for $x \geq a$. If $\int_{a}^{\infty} g(x) d x$ converges, then $\int_{a}^{\infty} f(x) d x$ also converges. In Problems $79-82,$ use this result to show that the given integral converges.$$\int_{1}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x$$

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