evaluate the improper integral or show that it diverges.\n int_{0}^{4} \frac{d x}{sqrt{16-x^{2}}} \nselect…

evaluate the improper integral or show that it diverges.\n int_{0}^{4} \frac{d x}{sqrt{16-x^{2}}} \nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. the improper integral converges and ( int_{0}^{4} \frac{d x}{sqrt{16-x^{2}}}= ) (type an exact answer, using ( pi ) as needed.)\nb. the improper integral diverges.

evaluate the improper integral or show that it diverges.\n int_{0}^{4} \frac{d x}{sqrt{16-x^{2}}} \nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. the improper integral converges and ( int_{0}^{4} \frac{d x}{sqrt{16-x^{2}}}= ) (type an exact answer, using ( pi ) as needed.)\nb. the improper integral diverges.

Answer

Explanation:

Step1: Recall the integral formula

The integral formula (\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{- 1}(\frac{x}{a})+C) (where (a > 0)). Here (a = 4), so (\int\frac{dx}{\sqrt{16 - x^{2}}}=\sin^{-1}(\frac{x}{4})+C).

Step2: Evaluate the definite - integral

We use the fundamental theorem of calculus (\int_{0}^{4}\frac{dx}{\sqrt{16 - x^{2}}}=\lim_{b\rightarrow4^{-}}\int_{0}^{b}\frac{dx}{\sqrt{16 - x^{2}}}). By the fundamental theorem of calculus (\int_{0}^{b}\frac{dx}{\sqrt{16 - x^{2}}}=\left[\sin^{-1}(\frac{x}{4})\right]{0}^{b}=\sin^{-1}(\frac{b}{4})-\sin^{-1}(0)). Since (\sin^{-1}(0) = 0) and (\lim{b\rightarrow4^{-}}\sin^{-1}(\frac{b}{4})=\sin^{-1}(1)). We know that (\sin^{-1}(1)=\frac{\pi}{2}) and (\sin^{-1}(0) = 0).

Answer:

A. The improper integral converges and (\int_{0}^{4}\frac{dx}{\sqrt{16 - x^{2}}}=\frac{\pi}{2})