Question

Determine whether the improper integral diverges or converges. evaluate integral if it converges: the integral from neg. infinity to pos. infinity of 4/(16+(x^2)) dx

Answer 1

the integrand is even so we can convert this to\[\int_{-\infty}^{\infty}\frac4{16+x^2}dx=2\int_{0}^{\infty}\frac4{16+x^2}dx\]\[=8\int_{0}^{\infty}\frac1{4^2+x^2}dx=8\lim_{n \rightarrow \infty}\int_{0}^{n}\frac1{4^2+x^2}dx\]which is going to give us an arctan thing...\[2\lim_{n \rightarrow \infty}\tan^{-1}(\frac x4)|_{0}^{n}=2\lim_{n \rightarrow \infty}\tan^{-1}(\frac n4)=\pi\]I hope I did that right!

Answer 2

oh good, wolfram agrees!

Answer 3

thanks!

Answer 4

Yup its correct.. lol i actually solved to verify to see that u already verified using wolphram :P