So I have the integral $\displaystyle \int - \frac{3y^2}{2}$ 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!
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Original integral $\displaystyle \int - \frac{3y^2}{2}$ over 0 to 4. Well I did $\displaystyle -\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24$ and 0 makes the whole second fraction 0 so we can ignore it... but apparently I'm wrong
Originally Posted by CleanSanchez Original integral $\displaystyle \int - \frac{3y^2}{2}$ over 0 to 4. Well I did $\displaystyle -\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24$ 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
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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