1. ## Help!

What is area of coloured surface from y=x^2 and y=x on [0,1] interval?

Image is on attachment.

Thank you.

2. Originally Posted by GreenMile
What is area of coloured surface from y=x^2 and y=x on [0,1] interval?

Image is on attachment.

Thank you.
You need to find the area BELOW the line $\displaystyle y = x$ on that interval, and subtract the area BELOW the curve [tex] y = x^2[tex] the on that interval.

The area below the curve is given by:

$\displaystyle \int_0^1 x^2 dx$.

The area below the line is given by:

$\displaystyle \int_0^1 xdx$.

Hence the area enclosed between them is:

$\displaystyle \int_0^1 x dx- \int_0^1 x^2 dx$

$\displaystyle \int_0^1 x - x^2 dx$

3. Thanks mush.

I have a test soon and I have to prepare.

I have another problem:

According to the graphic find f(-1)+f(0) ?

4. Originally Posted by GreenMile
Thanks mush.
http://www.mathhelpforum.com/math-he...l-posters.html

I have a test soon and I have to prepare.

I have another problem:
Do yourself and do us a favour and start a new thread if you have a new question, please!

According to the graphic find f(-1)+f(0) ?
Determine the distance of the graph from the x-axis at x = -1. It is 0.
Determine the distance of the graph from the x-axis at x = 0. It is 2.

Therefore: $\displaystyle f(-1) + f(0) = 0 + 2 = 2$