## jackkim4 2 years ago Find the interval [a,b] for which the value of the integral b s(2+x-x^2)dx a is maximum

1. jackkim4

help me :'(

2. UnkleRhaukus

is this your question ?$\int\limits_a^b(2+x-x^2)\text dx$

3. jackkim4

yes!!! sry first time here....

4. jackkim4

is the answer [-1, 2] or [-1,1/2] ??

5. helder_edwin

as @UnkleRhaukus pointed out, you have maximize $\large F(a,b)=\int_a^b(2+x-x^2)\,dx$

6. helder_edwin

which one gives u the greatest area?

7. helder_edwin

*value, sorry.

8. jackkim4

doesnt [-1, 1/2] wait no... it's [-1, 2]

9. UnkleRhaukus

what did you get when you integrated

10. jackkim4

27/6 ???

11. helder_edwin

which interval?

12. jackkim4

[-1,2]

13. UnkleRhaukus

$\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b$$=2(b-a)+\frac{(b-a)^2}2-\frac{(b-a)^3}{3}$ right?

14. jackkim4

yes!

15. UnkleRhaukus

actually i think i made a mistake

16. jackkim4

it's separated right

17. jackkim4

[2b+b^2/2-b^3/3]-[2a+2a^2/2-a^3/3]

18. UnkleRhaukus

$\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b$$\qquad\qquad=\left({2b}+\frac{b^2}2-\frac{b^3}3\right)-\left({2a}+\frac{a^2}2-\frac{a^3}3\right)$yeahs

19. UnkleRhaukus

so when is this maximum

20. jackkim4

at a= -1 b= 2 ! right?

21. sirm3d

Maximize F(b) by derivative, minimize F(a) by derivative.

22. UnkleRhaukus

$=b\left({2}+\frac{b}2-\frac{b^2}3\right)+a\left(-{2}-\frac{a}2+\frac{a^2}3\right)$

23. UnkleRhaukus

$=6b\left(12+3b-2b^2\right)+6a\left(2a^2-3a-12\right)$

24. sirm3d

$\large F \prime (b)=2+b-b^2=0$$\large \text{max at } b=2$$\large F \prime (a)=2+a-a^2=0$$\large \text{max at } a=-1$ interval: [-1, 2]

25. sirm3d

wow, my answer is among the choices given.

26. jackkim4

omg lol

27. sirm3d

typo mistake. it should be min at a = -1

28. jackkim4

yayyyyyy tytytyty! :')