Calculus

posted by .

The area bounded between the line y=x+4 and the quadratic function y=(x^2)-2x.
Hint: Draw the region and find the intersection of the two graphs. Add and subtract areas until the appropriate area is found.

I found the intersection points as (-1,3) and (4,8).
I'm not sure what to do with this. I think the answer is 125/6, but not sure.
Please show me the steps on how to get this answer if it's correct.

  • Calculus -

    Your intersection points are correct. Between those points, the y = x+4 curve is above the y = x^2 -2x curve.

    The area between the curves is the integral of x + 4 dx from x = -1 to 4, MINUS the integral of x^2 -2x between the same two x values. That equals

    x^2/2 + 4x @ x = 4 - (x^2/2 +4x) @ x = -1
    MINUS
    x^3/3 -x^2 @ x = 4 - (x^3/3 -x^2) @x = -1

  • Calculus -

    Your intersections are correct.
    the vertex of the parabola is at (1,-1)
    The zeros of the parabola are at(0,0) and at (2,0)
    Between x = 0 and x = 2, the parabola dips below the x axis

    we want the height of the line between y = x+4 and y=x^2-2x
    which is
    x+4 -x^2+2x
    or
    -x^2 + 3x + 4
    integrate that dx from -1 to +4

    -x^3/3 + 3x^2/2 + 4x

    at 4
    -64/3 + 24 + 16 = (-64+72+48)/3
    = 56/3

    at -1
    +1/3 +3/2 -4 = 2/6 + 9/6 - 24/6
    = -13/6
    so we want
    56/3 -(-13/6) = 112/6+13/6 = 125/6
    yep, agree 125/6

  • Calculus -

    Yeah. Thanks for showing me the steps

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calculus

    2. Sketch the region in the first quadrant that is bounded by the graphs of y=x^3, y=4X , and 2x+y-3=0 that lies below both straight lines. Find the area of this region. I can see the graph in front of me. which function do I subtract …
  2. Calculus AB

    y=6-x y=x^2 Find the area of the region by integrating with respect to x. Find the area of the region by integrating with respect to y. ------------------------------------ i got the intersection pts to be(-3,9)and (2,4)....i then …
  3. math

    Question-2; Consider the function f(x)= -cos3x -4sin3x. (a)Find the equation of the line normal to the graph of f(x) when x= pie/6 . (b)Find the x coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal. …
  4. math

    hilp meeee trying .... Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the …
  5. math

    hilp meeee trying .... Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the …
  6. math

    weloo Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the area enclosed by …
  7. Calculus

    1. Find the area of the region bounded by the curves and lines y=e^x sin e^x, x=0, y=0, and the curve's first positive intersection with the x-axis. 2. The area under the curve of y=1/x from x=a to x=5 is approximately 0.916 where …
  8. calculus

    1. Let S be the region in the first quadrant bounded by the graphs of y = e^(-x^2) y = 2x^2, and the y axis. y x = 2 , and the y-axis. a) Find the area of the region S. So when I integrate, is it from 0 to .5325?
  9. Calc

    Find the area of the region bounded by the curves y2 = x, y – 4 = x, y = –2, and y = 1. So far I have found that the area of the trapezoid which is 13.5. But for the other two areas I cannot find them. They could be: 27/2 22/3 …
  10. Calculus-Steve

    Find the area of the region between the graphs of f(x)=3x+8 and g(x)=x^2 + 2x+2 over [0,2]. I got 34/3. Calculus - Steve ∫[0,2] (x^2+2x+2) dx = 1/3 x^3 + x^2 + 2x [0,2] = 8/3 + 4 + 4 = 32/3 Why are you taking the antiderivative …

More Similar Questions