# int_(-oo)^oo 4/(16+x^2) dx Determine whether the integral diverges or converges. Evaluate the integral if it converges.

int_-infty^infty 4/(16+x^2)dx=

Divide both numerator and denominator by 16.

int_-infty^infty (1/4)/(1+x^2/16)dx=int_-infty^infty (1/4 dx)/(1+(x/4)^2)=

Substitute u=x/4 => du=1/4 dx. Bounds of integration remain the same because dividing infinity (or minus infinity) by four gives infinity.

int_-infty^infty (du)/(1+u^2)=arctan u|_-infty^infty=lim_(u to infty)arctan u-lim_(u to -infty)arctan u=

pi/2-(-pi/2)=pi

As we can see the integral converges and its value is equal to pi.

The image below shows the graph of the function and area under it corresponding to the integral. As we can see the graph of the function approaches x-axis (function converges to zero) as x goes to +-infty. We can also see that this convergence seams pretty "fast" which means that the integral should converge which we've already shown by calculating it.

## See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Images:
Image (1 of 1)
Approved by eNotes Editorial