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Help to understand a problem, integral

  1. Dec 19, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imageshack.us/a/img197/305/yhmn.jpg [Broken]

    What I want to do is calculate the area of the colored space. As you can see in the picture the function is -0.75x^2+3 for the graph.

    What I did is since the colored area reaches up to y5 and not y3 I took



    ∫(-0.75x^2+3)dx and then multiplied the answer by 2 to get the whole area
    as both sides are identical. I realise now that it can't be the correct way to do it though as when I type in "-0.75x^2+5" in the graph calculator the line at the x-axis changes. I would greatly appreciate help on how to solve this.

    OBS: This was a problem at a test I had a few days ago, can't imagining it going well. I won't know of the results yet but I'm just curious about this specific problem.
    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 19, 2013 #2
    It's a "-" sign inbetween the integrals. It gets messed up when I try to write it next to each other
  4. Dec 19, 2013 #3
    That won't work, graph the function ##y = -\frac{3}{4}x^2+5## to see why.

    Try extending the shaded area to create a rectangle. Can you figure out a way to use this new area (which is easily calculated) and the integral of the function to calculate the original shaded area?
  5. Dec 19, 2013 #4
    Ye, I know it didnt work :s I don't know how to calculate the new area if I just color in the rest of the rectangle.

    I would know how to calculate the function's area i.e the part I would color in, but the part that is already colored I don't know :s.
  6. Dec 19, 2013 #5
    Okay, think about it this way.

    A = The shaded area you are trying to find
    B = Area under the function and above the x-axis
    C = Area of the rectangle

    ? + ? = ?
    A = ?
  7. Dec 19, 2013 #6
    A = C-B?

    What I don't know is how to find the area of the rectangle. I know this is supposed to be super easy as it was one of the easier problems on the exam I had.
  8. Dec 19, 2013 #7
    That's correct for A! Area of a rectangle is just length times width. You don't need to do any fancy integrating (although you could if you wanted to).
  9. Dec 19, 2013 #8


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    Seriously? The area of a rectangle is "width times height". From your graph it looks like the height is 5 and the width is 4.

    You could also think of the upper limit of that rectangle as "y= 5" so the area of the rectangle is [itex]\int_{-2}^2 5 dx[/itex].

    The distance from the parabola up to that line is [itex]5- (-0.75x^2+ 3)= 2+ 0.75x^2[/itex] so the area between the line and the parabola is the integral of that.
  10. Dec 19, 2013 #9
    The integral ∫02(-0.75x^2+3dx) [0.75x^3/3+3x+c]20

    = -0.75*2^3/3+3*2+c - (0+c) = 4 +c-c = 4ae(area units) that is the area of the function (or half the function right? since it is x0->x2 and not x-2->x2)

    Then what I did was, just same thing, calculating he half of the rectangle. So:

    20(-0.75x^2+5)dx [-0.75x^3+5x+c]20

    = -0.75*2^3/3+5*2+c - (0+c) = 8+c-c = 8 ae. Then I took 8ae-4ae (half rectangle - half of the function area) which is 4ae, then I multiplied it by 2 and got 8ae for the entire colored space.

    But I think I just should've skipped the -0.75x^2 part huh. I get it now, damnit. Ye makes a lot of sense lol, I must be the stupidest person alive loooooooooooooooooooooool, the most obvious things are the hardest for me to see.
  11. Dec 19, 2013 #10
    Okay, your notation is getting a little sloppy. First of all, you don't need "ae" for area units. The integral computes an exact number. The ae just makes the computations more complicated because they can be confused for variables if variables are present in your integral. Tack on the units, if needed, at the end of the problem. Second of all, when computing definite integrals, you can drop the integrating constant because it cancels out. You still get the correct answer if you include it, but there is no point in doing so.

    I assume you meant to say ∫02(-0.75x^2+3dx) = [0.75x^3/3+3x+c]20. The computations would have been easier if you converted 0.75 to ##\frac{3}{4}## because ##\int (\frac{3}{4}x^2 + 3) dx = \frac{1}{4}x^3 + 3x + C##. Now you don't have to worry about multiplying 0.75 by 8 and divinding by 3.

    In any case, were you trying my approach or HallsofIvy's approach? The second integral is wrong for calculating the area of the rectangle. The correct integral would be ##\displaystyle\int_{-2}^2 5 dx##, as HoI noted, which equals 20. 20 minus your answer of 8 from the first integral yields the correct answer of 12.

    Doing it HallsofIvy's way yields ##\displaystyle\int_{-2}^2 (2 + \frac{3}{4}x^2) dx = [2x+\frac{1}{4}x^3]_{x = -2}^{x = 2}##
  12. Dec 19, 2013 #11
    The reason I use "ae" is because my teacher told me to always include it in my calculations.

    Ye, I realised it was wrong as noted, the second integral that is. I should've left out -3/4x^2 in that one to get the correct answer.

    Also I know C cancels each other out but my teacher told me always to use it in my calculations also.

    But yeah, this was a rly easy problem once I understood what I was doing wrong lol.
  13. Dec 20, 2013 #12


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    Yes, so there are several different ways you can do this:
    1) The area of the 4 by 5 rectangle is 4(5)= 20 and the area under the parabola is [itex]\int_{-2}^2 -(3/4)x^2+ 3 dx= \left[-(1/4)x^3+ 3x\right]_{-2}^2= (-(1/4)(8)+ 3(2))- (-(1/4)(-8)+ 3(-2))= (4)- (-4)= 8[/itex]. So the area between them is 20- 8= 12.

    2) The area of the rectangle is [itex]\int_{-2}^2 5 dx= \right[5x\left]_{-2}^2= 5(2)- 5(-2)= 10+ 10= 20. The area of the parabola is as before so the area between them is 20- 8= 12.

    3) The distance between the line, y= 5, and the parabola, [itex]y= -(3/4)x^2+ 3[/itex], is [itex]5- (-(3/4)x^2+ 3)= 2+ (3/4)x^2[/itex] so the area between them is [itex]\int_{-2}^2 2+ (3/4)x^2 dx= \left[2x+ (1/4)x^3\right]_{-2}^2= (2(2)+ (1/4)(8))- (2(-2)+ (1/4)(-8))= (4+ 2)- (-4- 2)= 6- (-6)= 12.

    Of course, we could have recognized that the graphs are symmetric about the y-axis so just done 0 to 2 and doubled that result.
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