Integration is driving me nuts!!!

**Johnaloa** · April 29th 2008, 02:37 AM

So the question is to do with Riemann sums.

Question: Use Riemann summs to find the exact value of the definite integral of

g(x) = 2 - (x^3)

on the interval [0,4]. The value found should not be the area under the graph on the interval. Why?

Note: You need to use equal subintervals and the sums will be slightly easier if you use the right hand endpoints.

Answer: I can do everything but (and this may sound silly) I dont know how to incorporate the consant '2' into the sums. Im thinking my answer should come to -56 right. But I keep gettin 64!

Can anyone fully show me how to do this?

Also Im guessing value found is not the area under the graph because its the signed area, its the area above the x-axis minus the area below the x-axis. Is this right?

P.S. Sorry about not showing any of my working. I'm tired and am going to go to sleep soon. Thanks in advance for your guys help.

**Mathstud28** · April 29th 2008, 02:54 AM

Originally Posted by Johnaloa

So the question is to do with Riemann sums.

Answer: I can do everything but (and this may sound silly) I dont know how to incorporate the consant '2' into the sums. .

Sorry I dont have time to answer your question if full...but for the constant just treat it seperately

$\int_a^{b}f(x)+2dx=\int_a^{b}f(x)dx+\int_a^{b}2dx$

**Danshader** · April 29th 2008, 02:57 AM

well i am not sure how you did your integration but here is how i would do mine:

$\int{g(x)} = \int^4 _0 {2- x^3} dx $
$ = 2x|^4_0 - \frac{1}{4}x^4|^4 _0 $
$ = (8-0) - \frac{1}{4}(4^4 - 0^4) $
$ = 8 - 64 $
$ =-56 $

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