anonymous 4 years ago 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

1. TuringTest

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!

2. TuringTest

oh good, wolfram agrees!

3. anonymous

thanks!

4. Akshay_Budhkar

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