## CalcDerp102 Group Title Im doing a problem at the moment and im stuck its a right riemann sum problem, equation and how far i got posted below one year ago one year ago

1. CalcDerp102 Group Title

$\int\limits_{3}^{0} (2x^2 + x +3) dx$ $\Delta x = -3/n$ $x_{k} = 3 + -3k/n$ $f(x_k) = (3+ -3k/n)^2 - (3 + -3k/n) +1$

2. CalcDerp102 Group Title

im not sure where to go from there

3. Algebraic! Group Title

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4. Algebraic! Group Title

I think you'd be better off using: $\huge \int\limits_{a}^{b} f(x) dx = -\int\limits_{b}^{a} f(x) dx$ at the start

5. Algebraic! Group Title

then $f(c _{k}) = 2*k^2 *\Delta x ^3 + k*\Delta x^2 + 3 \Delta x$

6. Algebraic! Group Title

and $S _{n} = 2 \Delta x ^3 \sum_{}^{} k^2 + \Delta x^2 \sum_{}^{} k +3 \Delta x$

7. CalcDerp102 Group Title

so if be better of making the original equation negative then using that formula to find the equation

8. Algebraic! Group Title

a bit easier, yes... that's probably why they gave it to you in the form that they did... so you could make the easy simplification

9. Algebraic! Group Title

you got it from here?

10. CalcDerp102 Group Title

yea i think so i think what confuses me the most is 3+(-3k/n)

11. Algebraic! Group Title

do what I did before you sub.s in 'delta x = b/n'

12. CalcDerp102 Group Title

$2 * (3+ \frac{ -3k }{ n }^2) * \frac{ -3 }{ n }^3 + (3+ \frac{ -3k }{ n }) * \frac{ -3 }{ n } + 3\frac{ -3 }{ n }$ is this what im looking for or do does k = 1

13. Algebraic! Group Title

start at x=0... move delta x to the right... what's the height of the rectangle you'd draw there?

14. Algebraic! Group Title

|dw:1354238714258:dw|

15. Algebraic! Group Title

height is f(delta x) width is delta x area is f(delta x) * delta x which is (2*(delta x)^2 + delta x +3))*delta x

16. Algebraic! Group Title

|dw:1354238854668:dw|

17. Algebraic! Group Title

next step: go another 'delta x' to the right height is 2*(2*delta x)^2 + 2*delta x +3 width is delta x area is (2*(2*delta x)^2 + 2*delta x +3)*delta x

18. Algebraic! Group Title

so the sum of areas is going to look like: (2*(k*delta x)^2 + k*delta x +3)*delta x

19. Algebraic! Group Title

2*k^2*(delta x)^3 + k*(delta x)^2 +3*delta x

20. Algebraic! Group Title

summing from k=1 to n

21. Algebraic! Group Title

that's the sum I wrote above

22. CalcDerp102 Group Title

oh alright, thanks i appreciate the help!

23. Algebraic! Group Title

does that make sense?

24. CalcDerp102 Group Title

Yeah it does, im trying to work through the problem. ill have an answer soon enough though

25. Algebraic! Group Title

still there?

26. Algebraic! Group Title

I didn't write the last term on the summation properly

27. Algebraic! Group Title

it should be:$\Delta x \sum_{k=1}^{n} 3$

28. Algebraic! Group Title

so$S _{n} = 2(\Delta x)^3\sum_{k=1}^{n} k^2 + (\Delta x)^2 \sum_{k=1}^{n} k +\Delta x \sum_{k=1}^{n} 3$

29. CalcDerp102 Group Title

oh ok, thats makes more sense now

30. CalcDerp102 Group Title

Its either 0 or -6 i think 0 because n = 3, am i correct?

31. Algebraic! Group Title

it's -63/2

32. Algebraic! Group Title

n is the number of divisions in the partition... you find the sum above (in terms of n) ... you sub.s in b/n for delta x and then you take the limit as n->infinity

33. Algebraic! Group Title

then use b=3 (the interval length) and finally, make the whole result negative ( due to the simplification we did at the very beginning)