Question

Which of the improper integrals in Exercises 63–68 converge and which diverge?∫ from 6 to ∞ of (1 / √(θ² + 1)) dθ

Accepted Answer

Identify the type of improper integral: Since the upper limit of integration is infinity, this is an improper integral of the form $$\int_{6}^{\infty} \frac{1}{\sqrt{\$$theta^{2}$$ + 1}} \, d	heta. $$Set up the integral as a limit: Rewrite the integral as $$\lim_{t 	o \infty} \int_{6}^{t} \frac{1}{\sqrt{\$$theta^{2}$$ + 1}} \, d	heta to $$handle the infinite upper bound. Find the antiderivative: Recognize that the integral of $$\frac{1}{\sqrt{\$$theta^{2}$$ + 1}}$$ with respect to $$	heta is$$ $$\$$sinh^{-1}$$(	heta) or $$equivalently $$\ln\left(	heta + \sqrt{\$$theta^{2}$$ + 1}ight). $$Evaluate the definite integral from 6 to $$t$$: Substitute the limits into the antiderivative to get $$\$$sinh^{-1}$$(t) - \$$sinh^{-1}$$(6) or$$ $$\ln\left(t + \sqrt{t$$^{2}$$ + 1}ight) - \ln\left(6 + \sqrt{36 + 1}ight). $$Take the limit as $$t$$ approaches infinity: Analyze $$\lim_{t 	o \infty} \$$sinh^{-1}$$(t) or$$ $$\lim_{t 	o \infty} \ln\left(t + \sqrt{t$$^{2}$$ + 1}ight) to $$determine if the integral converges (finite limit) or diverges (infinite limit).

Which of the improper integrals in Exercises 63–68 converge and which diverge?
∫ from 6 to ∞ of (1 / √(θ² + 1)) dθ

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