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Math Help - I'm having trouble finding the area under curves

  1. #1
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    I'm having trouble finding the area under curves

    Find the area of the region bounded by the parabola the tangent line to this parabola at (1, 1) that you found in (a), and the
    x-axis.


    The tangent line is y= 2x-1.

    Here's a graph of what it looks like..


    I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.
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  2. #2
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    Quote Originally Posted by cdlegendary View Post
    Find the area of the region bounded by the parabola the tangent line to this parabola at (1, 1) that you found in (a), and the
    x-axis.


    The tangent line is y= 2x-1.

    Here's a graph of what it looks like..


    I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.
    I get 1/3 for that integral too. Everything looks right to me if you want the area between the two curves from 0 to 1.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by cdlegendary View Post
    Find the area of the region bounded by the parabola the tangent line to this parabola at (1, 1) that you found in (a), and the
    x-axis.

    The tangent line is y= 2x-1.

    Here's a graph of what it looks like..


    I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.
    If we are to do this with respect to x, then we wil need two integrals.

    Note that the tangent line crosses the x axis at x=\frac{1}{2}

    And the two graphs intersext at x=1

    This implies that the Area in question may be found by

    A=\int_0^\frac{1}{2}x^2dx+\int_\frac{1}{2}^1[x^2-(2x-1)]dx
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  4. #4
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    Yeah, I really don't know whats wrong. The website I'm supposed to submit this to just isn't accepting that answer
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    If we are to do this with respect to x, then we wil need two integrals.

    Note that the tangent line crosses the x axis at x=\frac{1}{2}

    And the two graphs intersext at x=1

    This implies that the Area in question may be found by

    A=\int_0^\frac{1}{2}x^2dx+\int_\frac{1}{2}^1[x^2-(2x-1)]dx
    Ahh thanks! I see why it's like that now
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  6. #6
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    Actually, with that I got the same answer as before, 1/3.
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  7. #7
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    Alternatively, calculate the area under the curve from x=0 to x=1.
    Then subtract the area of the half-rectangle from x=0.5 to x=1

    \int_0^1{x^2}dx-0.5(0.5)=\frac{1^3}{3}-0.25=\frac{1}{3}-\frac{0.75}{3}=\frac{0.25}{3}=\frac{1}{12}

    The integral will include the area under the line and above the x-axis from x=0.5 to x=1, hence we must subtract the area under the line in order to find the area between the line and curve, bounded by the positive x and y axes, or take the route shown by Von Nemo.
    Last edited by Archie Meade; January 24th 2010 at 12:10 PM.
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  8. #8
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    Can you show us some intermediate work? Like what do you get for the two antiderivatives?

    The solution VonNemo19 gave doesn't add up to 1/3, so there must be a mistake somewhere if you are still getting 1/3.
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    Then subtract the area of the rectangle from x=0.5 to x=1
    Surely you mean to subtract the area of the triangle, which is (1/2)*0.5*1 = 0.25.

    I'm not sure if you arrived at the correct area by accident (the shape isn't a rectangle!) or just halved the height of the triangle in your head first.
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  10. #10
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    of course i meant a triangle
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by cdlegendary View Post
    Actually, with that I got the same answer as before, 1/3.
    Let's do it with respect to y, It'll be easier.

    Recall that

    A=\int_c^d(\text{right curve - left curve})dy

    when we want to find area between curves with respect to y.

    Step 1. Find limits of integration.

    We need to know over what interval y will move along. This is easy. From the context of the question (i.e find the are between curves from y=0 to y=1).

    So, we see that c=0 and d=1.

    Step 2 Determine which is the \mbox{right curve} and which is the \mbox{left curve}.

    If we were to draw a line which is parallel to the x-axis we would see that the curve y=x^2 would intersct this line to the left of where y=2x-1 would intersect said line. Therefore \mbox{rightcurve}:y=2x-1 and \mbox{leftcurve}:y=x^2.

    Step 3. Get every thing in terms of y

    y=x^2\mbox{ implies that }x=\sqrt{y}

    And

    y=2x-1\mbox{ implies that }x=\frac{y+1}{2}

    Step 4. Substitute the relevant information into the formula


    A=\int_c^d(\mbox{right curve - left curve})dy=\int_0^1\left(\frac{y+1}{2}-\sqrt{y}\right)dy

    Step 5. Evaluate

    If you need help with step 5, let me know.
    Last edited by VonNemo19; January 24th 2010 at 01:52 PM.
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  12. #12
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    Quote Originally Posted by VonNemo19 View Post
    Let's do it with respect to y, It'll be easier.

    Recall that

    A=\int_c^d(\text{right curve - left curve})dy

    when we want to find area between curves with respect to y.

    Step 1. Find limits of integration.

    We need to know over what interval y will move along. This is easy. From the context of the question (i.e find the are between curves from y=0 to y=1).

    So, we see that c=0 and d=1.

    Step 2 Determine which is the \mbox{right curve} and which is the \mbox{left curve}.

    If we were to draw a line which is parallel to the x-axis we would see that the curve y=x^2 would intersct this line to the left of where y=2x-1 would intersect said line. Therefore \mbox{rightcurve}:y=2x-1 and \mbox{leftcurve}:y=x^2.

    Step 3. Get every thing in terms of y

    y=x^2\mbox{ implies that }x=y^2

    And

    y=2x-1\mbox{ implies that }x=\frac{y+1}{2}

    Step 4. Substitute the relevant information into the formula


    A=\int_c^d(\mbox{right curve - left curve})dy=\int_0^1\left(\frac{y+1}{2}-y^2\right)dy

    Step 5. Evaluate

    If you need help with step 5, let me know.
    thanks, that worked out. Got the answer 1/12.
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