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Math Help - Calc2: Not sure where to begin...

  1. #1
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    Smile Calc2: Not sure where to begin...

    Hey, everyone!

    I have a test coming up next week and I'm practicing for it (so, the questions within are exclusively for my gain, not a grade) and there's a few problems on the book that I can't seem to get.

    Question #1: Sketch the region enclosed by \text{y}=8x+2 and \text{y}=5x^2. Decide whether to integrate with respect to \text{x} or \text{y}. Draw a typical approximating rectangle and label it's hight and width, then find the area of the region.

    What I have done: Okay, I drew the graph and I get (by equaling both functions) that the two functions intersect at (\frac{-11}{50} , \frac{6}{25}) and at (\frac{91}{50} , \frac{83}{5})

    Now, should I integrate from -11/50 to 0 and then from 0 to 91/50?

    Question #2: Use Calculus to find the area of the triangle with the given vertices: (0,0) , (4,1) , (-1,11)

    Here, I have no clue. I thought I could just find the midpoint between one of the segments and then the distance to the opposite vertex in order to get my height...?

    There was something about this type of exercise in the "Introductory Tutorial to Multiple Integration" thread, but since we haven't done that in class, I'm not sure I can do in the test.

    Sorry for the long read! I just need someone to point me in the right direction!
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  2. #2
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    Quote Originally Posted by Polyxendi View Post
    Hey, everyone!

    I have a test coming up next week and I'm practicing for it (so, the questions within are exclusively for my gain, not a grade) and there's a few problems on the book that I can't seem to get.

    Question #1: Sketch the region enclosed by \text{y}=8x+2 and \text{y}=5x^2. Decide whether to integrate with respect to \text{x} or \text{y}. Draw a typical approximating rectangle and label it's hight and width, then find the area of the region.

    What I have done: Okay, I drew the graph and I get (by equaling both functions) that the two functions intersect at (\frac{-11}{50} , \frac{6}{25}) and at (\frac{91}{50} , \frac{83}{5})
    That is not at all what I get. 5x^2= 8x+ 2 is equivalent to 5x^2- 8x- 2= 0 which has no rational roots. In any case, I see no reason for a double integral- especially since the problem says "decide whether to integrate with respect to x or y". That clearly does not imply integrating with respect to both! And integrating with respect to both x and y between constant limits would be integrating over a rectangle, not this region.

    Since the straight line is always above the parabola, you can just integrate 8x- 2- 5x^2 between the two x values. If you were to integrate with respect to y, you would have to change integrands at the lower point where the two graphs intersect.

    Now, should I integrate from -11/50 to 0 and then from 0 to 91/50?
    No, because (1) this is not a "double integral", (2) that would be integrating over a rectangle, not the region given, and (3) those numbers are wrong.

    Question #2: Use Calculus to find the area of the triangle with the given vertices: (0,0) , (4,1) , (-1,11)

    Here, I have no clue. I thought I could just find the midpoint between one of the segments and then the distance to the opposite vertex in order to get my height...?
    Well, the wouldn't "use Calculus", would it? You can "use Calculus" by treating this as an "area between graphs" problem. The straight line from (0, 0) to (-1, 11) is given by y= -11x and from (-1, 11) to (4, 1) by y= -2x+ 9. For x between -1 and 0, the triangle area is between those two graphs. The line from (0, 0) to (4, 1) is given by y= x/4. For x between 0 and 4, the triangle area is between y= -2x+ 9 (above) and y= x/4 (below).

    There was something about this type of exercise in the "Introductory Tutorial to Multiple Integration" thread, but since we haven't done that in class, I'm not sure I can do in the test.

    Sorry for the long read! I just need someone to point me in the right direction![/QUOTE]
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  3. #3
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    First of all, thank you for your time!

    I was actually wondering what were the points of intersection, so that I could have lower and upper limits in order to evaluate. This is what I have o far:

    \int(5x^2 - 8x - 2)dx

    \int5x^2dx - \int8xdx - \int2dx

    5\int{x^2}dx-8\int{x}dx-\int2dx

    \frac{5}{3}x^3 - \frac{8}{2}x^2 - 2x

    \frac{5}{3}x^3 - 4x^2 - 2x

    \frac{5}{3}x^3 - \frac{12}{3}x^2 - \frac{6}{3}x

    \frac{1}{3}x(5x^2-12x-6)

    Now, I need to evaluate, but I'm not sure from where to where.

    I tried the using the quadratic formula on 5x^2-8x-2 and I got

    \frac{-8\pm\sqrt{8^2+4(5)(2)}}{2(5)}

    \frac{-8\pm\sqrt{64+40}}{10}

    \frac{-8\pm\sqrt{24}}{10}

    Which give me:

    \frac{-8}{10}\pm\frac{2\sqrt6}{10}

    \frac{-4}{5}\pm\frac{\sqrt6}{5}

    And I ended up with \frac{-4+\sqrt6}{5} and \frac{-4-\sqrt6}{5}

    I tried solving them, numerically (i.e. giving value to \sqrt{6}, performing the operations, etc), but that's not the answer in the book...

    Thank you, though!
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  4. #4
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    Okay, so, I tried doing the integral through Wolfram and it gives me the upper and lower limit values as \frac{1}{5}(4+\sqrt{26}) and \frac{1}{5}(4-\sqrt{26}), respectively.

    How did it get those limits? When evaluated, it gives me the answer in the book....
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  5. #5
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    Quote Originally Posted by Polyxendi View Post
    First of all, thank you for your time!

    I was actually wondering what were the points of intersection, so that I could have lower and upper limits in order to evaluate. This is what I have o far:

    \int(5x^2 - 8x - 2)dx

    \int5x^2dx - \int8xdx - \int2dx

    5\int{x^2}dx-8\int{x}dx-\int2dx

    \frac{5}{3}x^3 - \frac{8}{2}x^2 - 2x

    \frac{5}{3}x^3 - 4x^2 - 2x

    \frac{5}{3}x^3 - \frac{12}{3}x^2 - \frac{6}{3}x

    \frac{1}{3}x(5x^2-12x-6)

    Now, I need to evaluate, but I'm not sure from where to where.

    I tried the using the quadratic formula on 5x^2-8x-2 and I got

    \frac{-8\pm\sqrt{8^2+4(5)(2)}}{2(5)}

    \frac{-8\pm\sqrt{64+40}}{10}

    \frac{-8\pm\sqrt{24}}{10}
    65+ 40 is NOT 24! It is 104= 4(26)

    Which give me:

    \frac{-8}{10}\pm\frac{2\sqrt6}{10}

    \frac{-4}{5}\pm\frac{\sqrt6}{5}

    And I ended up with \frac{-4+\sqrt6}{5} and \frac{-4-\sqrt6}{5}

    I tried solving them, numerically (i.e. giving value to \sqrt{6}, performing the operations, etc), but that's not the answer in the book...

    Thank you, though!
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  6. #6
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    HAHAHAHAHAHA!!!! I'm such an idiot!! lo!! Thank you SO much! Now I know that, for my test, I have to be careful with... arithmetic! THanks!
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