Definite integral problem

**CleanSanchez** · November 20th 2009, 03:10 AM

So I have the integral $\int - \frac{3y^2}{2}$ over 0 to 4.

Every computer software evaluates this as -32 which is the correct answer but I keep getting -24... what am I doing wrong? Any help is appreciated!

**Defunkt** · November 20th 2009, 03:19 AM

Show some work so we can tell you where you went wrong and what you need to do

**CleanSanchez** · November 20th 2009, 03:30 AM

Original integral $\int - \frac{3y^2}{2}$ over 0 to 4.

Well I did $-\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24$

and 0 makes the whole second fraction 0 so we can ignore it... but apparently I'm wrong

**Defunkt** · November 20th 2009, 03:34 AM

Originally Posted by CleanSanchez

Original integral $\int - \frac{3y^2}{2}$ over 0 to 4.

Well I did $-\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24$

and 0 makes the whole second fraction 0 so we can ignore it... but apparently I'm wrong

Did you not forget to integrate? :P

**CleanSanchez** · November 20th 2009, 03:42 AM

Ah yes, I was getting ahead of myself and thats not where my mistake was. It's part of a larger double integral problem and the bounds of x are 0 - 2 not 0 -4. Reversing double integrals can get confusing!

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