Question

calc help!!

Answer 1

i got 20.25

Answer 2

what are the bounds [a,b] for the integration?

Answer 3

x^3 = 9x

Answer 4

ill put up a graph...

Answer 5

looks like you have 2 regions, [-3,0] of integral of x^3 - integral of 9x, then
from [0,3] for integral of 9x- integral of x^3

Answer 6

I think they should be equal from symmetry.

Answer 7

so it would be 0?

Answer 8

for this i got 81/4

\[\int\limits_{-3}^{0}[x^3]dx - \int\limits_{-3}^{0}(9x)dx = \frac{ 81 }{ 4 }\]

Answer 9

now from 0 to 3...

Answer 10

oh so i was right? (:

Answer 11

\[\int\limits_{0}^{3}(9x)dx - \int\limits_{0}^{3}x^3dx = \frac{ 81 }{ 4 }\]

Answer 12

notice the top function is 9x in the second region, minus the bottom function x^3

Answer 13

They both come out to the same 81/4, the total area bounded by those two functions is, 2 times 81/4

Answer 14

are you good with evaluating those integrals, i just put the answers up there...

Answer 15

you were right, but you had to double your answer to account for both regions.

Answer 16

so my final answer would be 40.50? @DanJS

Answer 17

i would just leave it as 81/2

Answer 18

wait im confused. didnt you just say i have to double it??

Answer 19

i have to put it into decimal form

Answer 20

yes, each integral came out to 81/4, double that is 81/2

Answer 21

ohh got it!

Answer 22

40.50

Answer 23

yeah