Posted by **Anonymous** on Sunday, October 21, 2007 at 11:25am.

Find the area of the region in the first quadrant enclosed by the coordinate axes and graph of x^3+y^3=1.

- calculus -
**bobpursley**, Sunday, October 21, 2007 at 11:39am
Isnt the height of the function (1-x^3)^1/3 ?

so integrate INT (height)dx from x=0to1

Having a graph(graphing calc) will help understand.

- calculus -
**Anonymous**, Sunday, October 21, 2007 at 11:50am
Is it possible to do it by hand? how would I do it by a graphing calculator?

- calculus -
**bobpursley**, Sunday, October 21, 2007 at 12:53pm
The graphing calculator is to graph the function, so you understand what your integral is supposed to do.

Sure, you can graph it by hand.

- calculus -
**drwls**, Sunday, October 21, 2007 at 12:55pm
I cannot find a closed-form solution for the integral in my Tables of Integrals, but the integral can be obtained quite accurately and easily using Simpson's-rule numerical integration. At x values of 0,0.25, 0.5. 0.75 and 1, the values of the function are:

1, 0.995, 0.957, 0.833, and 0

Using the Simpson's Rule formula, I get 0.707 for the integral. Higher accuracy can be obtained by breaking up the 0 to 1 range of integration into larger number of slices. I used four (five data points).

- calculus -
**Anonymous**, Sunday, October 21, 2007 at 2:50pm
I haven't learned the Simpson's Rule formula yet.

- calculus -
**Count Iblis**, Sunday, October 21, 2007 at 2:53pm
Substitute x = y^1/3 to express it in terms of the Beta function. The integral can then be expressed as:

1/3 Gamma(4/3)Gamma(1/3)/Gamma(5/3) = 0.883319375...

Gamma(x) = (x-1)! and most calculators will evaluate it correctly for fractional values of x.

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