Asked by cw
Find the exact area of the region bounded by the -axis and the graph of (x^3)-x
Answers
Answered by
MathMate
The function x^3-x is an odd function, and so it makes an area with the x-axis twice. Not sure if it is the sum of both areas that is needed.
Write
f(x)=x³-x=x(x²-1)
Therefore f(x) cuts the x-axis at -1, 0, and 1.
The area to the right of the y-axis can be found by the integral:
A = ∫f(x)dx from 0 to 1
=∫ (x³-x)dx
=[ x^4/4 - x²/2 ] from 0 to 1
=[0- 1/2]
=-1/2
The integral is negative because the area is below the x-axis. The area to the left of the y-axis has the same area.
See:
http://img121.imageshack.us/img121/8791/1296841065.png
Write
f(x)=x³-x=x(x²-1)
Therefore f(x) cuts the x-axis at -1, 0, and 1.
The area to the right of the y-axis can be found by the integral:
A = ∫f(x)dx from 0 to 1
=∫ (x³-x)dx
=[ x^4/4 - x²/2 ] from 0 to 1
=[0- 1/2]
=-1/2
The integral is negative because the area is below the x-axis. The area to the left of the y-axis has the same area.
See:
http://img121.imageshack.us/img121/8791/1296841065.png
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