# Thread: Sketch and Find the area of the region determined by the intersections of the curves.

1. ## Sketch and Find the area of the region determined by the intersections of the curves.

Sketch and Find the area of the region determined by the intersections of the curves.

y= (2/(x2+1)), y=(absolute value of) x

I do not know how to proceed. The book I am using is of no help. Perhaps a push in the right direction is in order?

2. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Well....I have given this one a go, and here is what I have come up with......

My sketch.....

Here is my worked out problem....

If you can, let me know if I am right (or wrong)...Thanks!

3. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

That's not entirely correct. You need to split it into two separate integrals because the function on the right side is x and the function on the left side is -x.

Technically the below is correct:

$\int_{-1}^{1} \frac{2}{1+x^2} - \left | x \right | dx$

But that's not what you did - you dropped the absolute value.

4. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Oh, I thought that I did seperate it into two different integrals...and I also thought the absolute value of x is simply x....

5. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

I would use the symmetry of the two even functions to state the area A of the region is:

$A=2\int_0^1\frac{2}{x^2+1}-x\,dx=\pi-1$

6. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Originally Posted by MarkFL2
I would use the symmetry of the two even functions to state the area A of the region is:

$A=2\int_0^1\frac{2}{x^2+1}-x\,dx=\pi-1$
where did you come up with the $\pi-1$

......and also why are you evaluating the integral from 0 to 1......should it not be from -1 to 1 as I did in my example?

Thanks in advance

7. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Originally Posted by JDS
where did you come up with the $\pi-1$

$2\int_0^1\frac{2}{x^2+1}-x\,dx=\mathop {\left. {4\arctan (x) - x} \right|}\nolimits_{x = 0}^{x = 1}=?$

If $g$ is an even function then $\int_{ - a}^a {g(x)dx} = 2\int_0^a {g(x)dx}$

8. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Originally Posted by JDS
where did you come up with the $\pi-1$

......and also why are you evaluating the integral from 0 to 1......should it not be from -1 to 1 as I did in my example?

Thanks in advance
You have already been told that in order to evaluate this area, you need to do TWO integrals, because the absolute value function is actually a HYBRID function, \displaystyle \begin{align*} |x| = \begin{cases} \phantom{-} x \textrm{ if } x \geq 0 \\ -x \textrm{ if } x < 0 \end{cases} \end{align*}, so you will need to perform an integral for EACH of those cases.

9. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Originally Posted by Prove It
You have already been told that in order to evaluate this area, you need to do TWO integrals, because the absolute value function is actually a HYBRID function, \displaystyle \begin{align*} |x| = \begin{cases} \phantom{-} x \textrm{ if } x \geq 0 \\ -x \textrm{ if } x < 0 \end{cases} \end{align*}, so you will need to perform an integral for EACH of those cases.
Thanks for your assistance...(and everyone thus far)... but I am not sure how to set it up to find those two different integrals, any advice?

10. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Hint: |x| = -x when x < 0 and |x| = x when 0 ≤ x.

11. ## Re: Sketch and Find the area of the region determined by the intersections of the cur

Thanks everyone, here is my solution.

NOTE: I wanted to make sure and give much thanks to Jacek! Jacek has been tutoring me in his spare time and is an Excellent Teacher!!! Thank you Jacek!