Thread: Definite integral

Evaluate each Definite integral.
∫ lower limit 1, upper limit 8. 3-y^(1/3) / y^(2/3) ; dy
I did
∫(3*y^(-2/3) - y^(-1/3)) dy from 1 to 8

9*y^(1/3) - 3/2*y^(2/3) eval. from 1 to 8

= 9(2) - 6 - 9 + 3/2

= 3 + 3/2 = 9/2
Is that right? Thanks

I cant really make out properly what the integral is. If you have to integrate 3-(y^1/3)/(y^2/3), the answer will be different. Can you type the problem out properly again. With the proper parentheses?

The answer is correct with your evaluation though.

Originally Posted by rowdy3

Evaluate each Definite integral.
∫ lower limit 1, upper limit 8. 3-y^(1/3) / y^(2/3) ; dy
I did
∫(3*y^(-2/3) - y^(-1/3)) dy from 1 to 8

9*y^(1/3) - 3/2*y^(2/3) eval. from 1 to 8

= 9(2) - 6 - 9 + 3/2

= 3 + 3/2 = 9/2
Is that right? Thanks

You can check your answer here: integrate (3 - y^(1/3))/y^(2/ 3) from y = 1 to y = 8 - Wolfram|Alpha

Originally Posted by Maanasi

I cant really make out properly what the integral is. If you have to integrate 3-(y^1/3)/(y^2/3), the answer will be different. Can you type the problem out properly again. With the proper parentheses?

The answer is correct with your evaluation though.

∫ lower limit 1, upper limit 8. (3-y^(1/3)) / y^(2/3) ; dy
I'll try to scan it.

Originally Posted by rowdy3

∫ lower limit 1, upper limit 8. (3-y^(1/3)) / y^(2/3) ; dy
I'll try to scan it.

In the link in my reply:

1. Your integral is clear. (In fact, with a small amount of effort it is clear from your OP as well).

2. Your question is answered.