area beneath a curve

[insanerp]

Hi all,

y=cuberoot(x) 0<equal to(x)<equal to 8

I know the area is roughly ~8 but how do I get the correct answer the x,y point I know are [0,0],[1,1],and[8,2]
Thanks,
Michele

 

[tkhunny]

Have you any specific direction?

Areas of Trepezoids might produce a reasonable approximation.

[(0+1)/2] * 1 = (1/2)*1 = 1/2
[(1+2)/2] * 7 = (3/2)*7 = 21/2
21/2 + 1/2 = 22/2 = 11 and that should be short. I'm not liking your '8' very much.

Are you counting only the WHOLE squares? Those partial ones can add up!

Note: The complete, exact answer is 12.

 

[sgtpepper]

You can solve for the area under curves with integrals. I don't know what level math you are in, but if the word intergrals rings a bell, then you should know what to do. Since you are given the limits of the area, you would use a definite intergral.
I don't know how to show the intergral notation online, so I'll just tell you that the intergral of the cube root of x is (3/4)*x^(4/3) and then to solve for the given limits you would plug the upper limit (8) into the intergral, then subtract the intergral with the lower limit (0) plugged in:

http://tinyurl.com/2ojaza

I get 12 also, like the person above me, I just thought you'd like to know how to get the exact answer!