Calculus (Area Between Curves)

posted by .

Find the area of the region bounded by the curves y^2=x, y-4=x, y=-2 and y=1
(Hint: You'll definitely have to sketch this one on paper first.) You get:
a.) 27/2
b.) 22/3
c.) 33/2
d.) 34/3
e.) 14

  • Calculus (Area Between Curves) -

    Since y^2 = x is not even defined left of the y-axis, the region is

    x<3: a triangle bounded by y=x+4, x=-3, y=-2
    area = 3*3/2 = 9/2

    -3<x<0: a rectangle, area 3*3=9

    0<=x<=1: a curved triangular area
    area = Int(1-√x)dx [0,1]
    = (x - 2/3 x√x)[0,1]
    = (1 - 2/3) = 1/3

    Total: 9/2 + 9 + 1/3 = 14 - 1/6

    Almost (e), but not quite. Typo somewhere?

  • Calculus (Area Between Curves) -

    find the definite integral y^2 - y + 4 evaluated from -2 to 1. I go 16.5 sq units Choice c

  • Calculus (Area Between Curves) -

    33/2 is correct.

  • Calculus (Area Between Curves) -

    Thank you Nade for making this correction! I went back and graphed the functions with pencil to see if I could figure a rough estimate for the area. I found that these function make an odd trapezoid shape. The area of the solid trapezoid portion of the graph is 15 and then there is a small portion on the other side of the y-axis. So, I, too, figured that 33/2 would be the most reasonable answer. Thank you both!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus (Area Between Curves)

    Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859 Based on my calculations, I …
  2. Calculus (Area Between Curves)

    Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859
  3. calculus

    Consider the curves y = x^2and y = mx, where m is some positive constant. No matter what positive constant m is, the two curves enclose a region in the first quadrant.Without using a calculator, find the positive constant m such that …
  4. Calculus

    We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please?
  5. Calculus

    We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please?
  6. Calculus-Area between curves

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even …
  7. Calculus Area between curves

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2-x=1
  8. CALCULUS

    Sketch the region enclosed by the given curves. y = 4/X y = 16x, y = 1X/16 x > 0 and the area between the curves
  9. Calculus

    Find the area of the region bounded by the curves y=12-x^2 and y=x^2-6. Hint:The answer should be a whole number.
  10. Math

    The curves y=sinx and y=cosx intersects twice on the interval (0,2pi). Find the area of the region bounded by the two curves between the points of intersection.

More Similar Questions