Question

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.∫₂^∞ (1 / (x√x)) dx

Accepted Answer

Rewrite the integral to a simpler form by expressing the integrand with exponents: $$\$$int_2$$^{\infty} \frac{1}{x \sqrt{x}} \, dx = \$$int_2$$^{\infty} x$$^{-1}$$ \cdot x$$^{-\frac{1}$${2}} \, dx = \$$int_2$$^{\infty} x$$^{-\frac{3}$${2}} \, dx. $$Recognize that this is an improper integral with an infinite upper limit, so express it as a limit: $$\lim_{t 	o \infty} \$$int_2$$^{t} x$$^{-\frac{3}$${2}} \, dx. $$Find the antiderivative of the integrand $$x^{-\frac{3}$${2}}$$. Recall the power rule for integration: $$\int x^n \, dx = \frac{$$x^{n+1}$$}{n+1} + C$$, where n$$ 
eq -1$$. Here, n$$ = -\frac{3}{2}$$, so the antiderivative is $$\frac{$$x^{-\frac{1}$${2}}}{-\frac{1}{2}} + C = -2 $$x^{-\frac{1}$${2}} + C$$. Evaluate the definite integral from 2 to t$ using the antiderivative: $$\left[-2 $$x^{-\frac{1}$${2}} ight]_$$2^{t} = -2$$ $$t^{-\frac{1}$${2}} + 2 \cdot $$2^{-\frac{1}$${2}}$$. Take the limit as t$ approaches infinity: $$\lim_{t 	o \infty} \left(-2 $$t^{-\frac{1}$${2}} + 2 \cdot $$2^{-\frac{1}$${2}}ight)$$. Determine if this limit exists (finite) to conclude whether the integral converges or diverges.

In Exercises 69–80, determine whether the improper integral converges or diverges. If it converges, evaluate the integral.
∫₂^∞ (1 / (x√x)) dx

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Improper Integrals

Convergence and Divergence of Integrals

Integration of Power Functions