# Area Between Three Curves

Okay so I'm not just looking for answer but I really need help conceptualizing which curve when using more than 2 belongs as the "top curve".

The problem I'm working with right now is

$y=x^2$, $y=2x-1$ ,$y=0$

I've graphed them out and found the points of intersection to be $(0,0), (1/2,0),$ and $(1,1).$

now when I take the integral I understand that the answer will be two integrals summed together. The first limit will be from $\int^.5_0$ (fractions don't appear to play well with the knowledge I have of formatting) and the second will be $\int^1_.5$.

What I don't under stand is why the solution isn't

$\int^.5_0 2x-1 -( 0)dx + \int^1_.5 x^2 -(2x -1)dx$

which once solved would give you $23/24$

Checking my work with Wolfram Alpha says it should be

$\int^.5_0 0 - (x^2)dx + \int^1_.5 2x -1 - (x^2)dx$

which equals $-1/12$

but $x^2$ is the upper boundary throughout the entire graph. So I don't understand why Wolfram Alpha subtracts it twice or why it is included in the integral of $\int^.5_0$ because that intersection point is of $y=0$ and $y=2x-0$ I'm going to start finding the area of shells and washers and discs so I really want to understand this.

• Well, it's possible that you entered something wrong in Wolfram Alpha. What was your input? Can you post a link? – Cameron Buie Oct 24 '15 at 3:42
• wolframalpha.com/input/… – Wendel ZEnmaster Oct 24 '15 at 3:45
• I also get $\frac{1}{12}$. You can do the integral of the quadratic from $(0,1)$ and then subtract the area of the triangle with base $.5$ and height $1$. I believe the answer is indeed $\frac{1}{12}$. Note: plus one-twelth. – Cehhiro Oct 24 '15 at 3:50

From what I can tell, both answers are incorrect. Wolfram Alpha's is the exact opposite of what it should be, for some reason. (The input order matters, apparently.) Since $x^2$ is always the upper boundary, it should instead be $$\int_{0}^{\frac{1}{2}}(x^2-0)\,dx+\int_{\frac{1}{2}}^{1}(x^2-(2x-1))\,dx.$$

So, you were correct on the second interval, but incorrect on the first, since you were taking an area there that was not bounded by the three functions (only by two, and the line $x=0$).