Integration: Area between two parabolas on a given interval

[burritth]

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x^2 − 5x + 2 and y = −x^2 + 5x − 6 for x in [0, 4]

 

[MarkFL]

Hello and welcome to MHB, burritth! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

I would begin by finding the $x$-coordinates of the points of intersection first, and at these points we have the $y$-coordinates being equal, so if we equate the two given functions, and solve for $x$, we will know where they intersect. What do you find?

 

[Prove It]

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x^2 − 5x + 2 and y = −x^2 + 5x − 6 for x in [0, 4]

So have you at least drawn the graphs?

 

[MarkFL]

Here's why I wouldn't bother with the graphing...we are given two parabolic curves, and we can see (by inspecting the sign of the coefficient of the squared terms) that:

$$f_1(x)=x^2-5x+2$$

opens upwards while:

$$f_2(x)=-x^2+5x-6$$

opens downwards. So, if we find two real roots for $f_1=f_2$, which we will call $x_1,\,x_2$ where $x_1<x_2$, then we know:

$$f_1>f_2$$ on $$(-\infty,x_1)\,\cup\,(x_2,\infty)$$

$$f_2>f_1$$ on $$(x_1,x_2)$$

So, this is why I was suggesting to begin with finding $x_1$ and $x_2$. :)

 

[karush]

Here is desmos screen shot