Question

Find the area of the region bounded above by y = 2x + 2 and below by y = x^2 - 1.

Answer 1

A. 23/3
B. 4.5
C. 32/3
D. 64/3

Answer 2

@Jhannybean

Answer 3

So first you need to identify what function is above and what is below. The higher bound and lower bound.

Answer 4

ok

Answer 5

To find the limits of the area we are integrating, we analyze the x-values that are bounded between these two functions, therefore we're trying to find the area between -1 and 1.

Answer 6

we're integrating from top to bottom, so we'll be taking "dx"

Answer 7

i already did i got 64/3 but i'm not sure

Answer 8

I think i have my limits off... is it from -1 to 3?

Answer 9

I can't quite tell where the parabola is going to..

Answer 10

ok

Answer 11

Oh..hm... maybe it's \[\large A= \int\limits_{-1}^{3}[(2x+2)-(x^2-1)]dx\]\[\large A=\int\limits_{-1}^{3}(-x^2+2x+3)dx\]\[\large A= [-\frac{ x^3 }{ 3 }+x^2+3x]\] evaluated from -1 to 3 \[\large A= [-\frac{ (-1)^3 }{ 3 }+(-1)^2+3(-1)]-[-\frac{ (3)^3 }{ 3 }+(3)^2+3(3)]\]\[\large A= [\frac{ 1 }{ 3 }+1-3]-[-9+9+9]\]\[A= \frac{ 13 }{ 3 }-(-9)\]\[A= \frac{ 40 }{ 3 }\]

Answer 12

AGH, it's the OTHER way around,-__- i'm sorry!!

Answer 13

its okay

Answer 14

but look at my multiple choices there is no 59/3 :/

Answer 15

A. 23/3
B. 4.5
C. 32/3
D. 64/3

Answer 16

... i'm not doing this right at all am i, lol. it's not 18,it's 9. Shall i call someone else in...

Answer 17

A=[−9+9+9]−[1/3+1−3]
A= 9-(-5/3) = 32/3 ************************

Answer 18

after the third try :| goodness. Shame on me, lol.

Answer 19

lol

Answer 20

so its C?

Answer 21

mmhmm

Answer 22

mhmm yes or mhmm no lol

Answer 23

yes :D lol

Answer 24

lol thanks :)

Answer 25

no problem,sorry about the confusion at first :\