Question

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

Answer 1

help me :'(

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

which one gives u the greatest area?

Answer 7

*value, sorry.

Answer 8

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

Answer 9

what did you get when you integrated

Answer 10

27/6 ???

Answer 11

which interval?

Answer 12

[-1,2]

Answer 13

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

Answer 14

yes!

Answer 15

actually i think i made a mistake

Answer 16

it's separated right

Answer 17

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

Answer 18

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

Answer 19

so when is this maximum

Answer 20

at a= -1 b= 2 ! right?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

wow, my answer is among the choices given.

Answer 26

omg lol

Answer 27

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

Answer 28

yayyyyyy tytytyty! :')