# Thread: Area between two functions

1. ## Area between two functions

How do I find the area of the region bounded by the graphs y = root x, y= x-2 and y= 1

I did the integral of root x minus 1 from 1 to 3 plus the integral of root x minus x plus 1 from 3 to 4. Apparently this is wrong, how should it be set up as?

I apologize for not entering it in a proper math format, I have no idea how to do so.

2. You MUST have some familiarity with basic functions.

Find the intersections:

(3,1), (1,1), (4,2)

Decide on applicable regions.

$\sqrt{x} > 1\;on\;(1,3)$

The it switches to

$\sqrt{x} > x-2\;on\;(3,4)$

And you are almost done.

$\int_{1}^{3} \sqrt{x} - 1\;dx + \int_{3}^{4} \sqrt{x} - (x-2)\;dx$

Alternatively, and I'll let you explore where this comes from, it's a little easier on the y-axis:

$\int_{1}^{2} (y+2) - y^{2}\;dy$

3. Find the area bounded by the equations bellow.

$y=\sqrt{x}$
$y=x-2$
$y=1$

$\int^{1}_{0}\sqrt{x}dx=\frac{1^{1.5}}{1.5}=\frac{2 }{3}=.\overline{66}$

$\int^{3}_{1}1d=3(1)-1(1)=2$
$\int^{3}_{2}x-2dx=\int^{1}_{0}xdx=\frac{1}{2}$

$2.\overline{66}-\frac{1}{2}=2.1\overline{66}$

edit: sorry this is wrong... forgive my stupidity

4. Please rethink that, Integral. Just exactly what region is bounded by those equations? The bounds do not specify either the x-axis or the y-axis.