Calculus

posted by .

Find the exact area below the curve (1-x)*x^9 and above the x axis

  • Calculus -

    Were you given (a,b) values?

    From x = ? to ?

  • Calculus -

    First you define the function as
    f(x)=(1-x)*x^9

    Inspection of its 10 factors should indicate that the function crosses the x-axis at two points (0,0), (0,1).

    The leading coefficient is -x^10, which means that the major part of the function is concave downwards, and is above the x-axis only between x=0 and 1.

    Check: f(0.5)=(1-0.5)*0.5^9 > 0

    The area sought is thus between the limits of x=0 and x=1.

    The area below a curve f(x) is
    ∫f(x)dx between the limits of integration (0 to 1).

    The function can be split up into a polynomial with two terms, and is thus easy to integrate.

    Inspection of the graph between 0 and 1 and an approximate calculation of the area shows that the area should be in the order of 0.01. Post your answer for a check if you wish.

    Here's a graph of the function between 0 and 1.

    http://img411.imageshack.us/img411/1397/1296510221.png

  • Calculus -

    I just looked at the graph.

    You want the area from x = 0 to x = 1.

    | = integrate symbol
    | x^9(1 - x)
    | x^9 - x^10

    Then plug in x, from 0 to 1.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus 12th grade.

    Find the area lying above the x axis and below the parabolic curve y= 4x -x^2 a.8 b. 8 1/3 c. 10 2/3 d. 16 B or C not sure?
  2. Calculus I

    Find the exact area of the surface obtained by rotating the curve about the x-axis. y = sqrt(1+5x) from 1<x<7
  3. Calculus I

    Find the exact area of the surface obtained by rotating the curve about the x-axis. y = sqrt(1+5x) from 1<x<7
  4. calculus

    Find area of the region under the curve y = 5 x^3 - 9 and above the x-axis, for 3¡Ü x ¡Ü 6. area = ?
  5. Calculus

    Find the exact area below the curve y=x^3(4-x)and above the x-axis
  6. Calculus

    Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the …
  7. calculus

    I'm having trouble on this question: Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.
  8. Calc

    R is the region below the curve y=x and above the z-axis from x=0 to x=b, where b is a positive constant. S is the region below the curve y=cos(x) and above the x-axis from x=0 to x=b. For what value of b is the area of R equal to …
  9. calculus

    Find the exact area of the surface obtained by rotating the curve about the x-axis. y=((x^3)/4)+(1/3X), 1/2≤X≤1
  10. calculus

    Find the exact area of the surface obtained by rotating the given curve about the x-axis. x=t^3, y=t^2, 0 ≤ t ≤ 1

More Similar Questions