# Thread: Area Between Two Curves - quick check and simplifying answers

1. ## Area Between Two Curves - quick check and simplifying answers

Hi all,

To make things easier, I took a picture of my work for these two problems. For the first one, I want to know how to simplify them to $\displaystyle \sqrt 8$ which if I'm not mistaken is the answer. For the second problem, I would also like to know how to simplify my answer, and perhaps a verification that my un-simplified answer is at least correct up to that point.

Thanks for any help!

2. ## Re: Area Between Two Curves - quick check and simplifying answers

Your working for Q.1 looks alright. Note that

\displaystyle \displaystyle \begin{align*} \frac{\sqrt{2} + \sqrt{2}}{2} + \frac{\sqrt{2} + \sqrt{2}}{2} &= \frac{2\sqrt{2}}{2} + \frac{2\sqrt{2}}{2} \\ &= \sqrt{2} + \sqrt{2} \\ &= 2\sqrt{2} \\ &= \sqrt{4} \cdot \sqrt{2} \\ &= \sqrt{ 4 \cdot 2 } \\ &= \sqrt{8} \end{align*}

Of course, \displaystyle \displaystyle \begin{align*} 2\sqrt{2} \end{align*} is considered to be a simpler form than \displaystyle \displaystyle \begin{align*} \sqrt{8} \end{align*}.

3. ## Re: Area Between Two Curves - quick check and simplifying answers

Thank you! That's very helpful, especially the step by step. I'm extremely rusty on simplifying the addition of radicals, so it's always good to see an explanation rather than just plug it in on your calculator and realize it's the same as $\displaystyle \sqrt 8$.

4. ## Re: Area Between Two Curves - quick check and simplifying answers

Another question about one of these problems (also, if anyone could still verify #2, that'd be awesome!)

Question is to find the area between $\displaystyle y= \frac {2}{x^2+1}$ and $\displaystyle |x|$.

I used symmetry to set the integral from 0 to 1, took the 2 out to make the integrand $\displaystyle arctan$, and realized that between 0 and 1, the absolute value of x is 1/2.

After applying the FTC, I got the answer of $\displaystyle \pi$. However, after using the fnInt() feature on my calculator, I got $\displaystyle \pi -1$. Not sure which is right or wrong, because after calculating everything I got two similar but different answers.

5. ## Re: Area Between Two Curves - quick check and simplifying answers

By observation, it's clear that you need to integrate between -1 and 1, but you are correct, by symmetry you can integrate between 0 and 1 and double.

\displaystyle \displaystyle \begin{align*} \int_{-1}^1{ \frac{2}{x^2 + 1} - |x| \, dx } &= 2\int_0^1{ \frac{2}{x^2 + 1} - x\, dx } \textrm{ since } |x| = x \textrm{ when } x \geq 0 \\ &= 2 \left[ 2\arctan{(x)} - \frac{x^2}{2} \right]_0^1 \\ &= 2 \left\{ \left[ 2\arctan{(1)} - \frac{1^2}{2} \right] - \left[ 2\arctan{(0)} - \frac{0^2}{2} \right] \right\} \\ &= 2 \left[ 2\left( \frac{\pi}{4} \right) - \frac{1}{2} \right] \\ &= 2\left( \frac{\pi}{2} - \frac{1}{2} \right) \\ &= \pi - 1 \end{align*}

6. ## Re: Area Between Two Curves - quick check and simplifying answers

Thank you once again! I assume my error was that I took the 2 from $\displaystyle \frac {2}{x^2+1}$ outside and made it 4, which I guess led to some bad calculations that cancelled the -1 from $\displaystyle \pi -1$. Once again you've been a huge help!