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

Image is on attachment.

Thank you.

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- Jan 14th 2009, 05:07 AM #1

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- Jan 14th 2009, 05:19 AM #2

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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 $

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

- Jan 14th 2009, 10:50 AM #3

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- Jan 14th 2009, 11:22 AM #4
http://www.mathhelpforum.com/math-he...l-posters.html

I have a test soon and I have to prepare.

I have another problem:**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 = 0. It is 2.

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