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  1. #1
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    Help!

    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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  2. #2
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    Quote Originally Posted by GreenMile View Post
    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  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:

     \int_0^1 x^2 dx.

    The area below the line is given by:

     \int_0^1 xdx.

    Hence the area enclosed between them is:

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

     \int_0^1 x -  x^2 dx
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  3. #3
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    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) ?
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  4. #4
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    Quote Originally Posted by GreenMile View Post
    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: f(-1) + f(0) = 0 + 2 = 2
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