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

Isnt the height of the function (1x^3)^1/3 ?
so integrate INT (height)dx from x=0to1
Having a graph(graphing calc) will help understand. 
Is it possible to do it by hand? how would I do it by a graphing calculator?

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. 
I cannot find a closedform solution for the integral in my Tables of Integrals, but the integral can be obtained quite accurately and easily using Simpson'srule 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). 
I haven't learned the Simpson's Rule formula yet.

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) = (x1)! and most calculators will evaluate it correctly for fractional values of x.