Question

sketch the graph bounded by the graphs of the given equation and find the area of that region

y=x^3+1
y=x-1

x=-1 x=1

this is area between curves

Answer 1

for starters; add +2 to it all to get it above the xaxis;

Answer 2

http://lj010.k12.sd.us/graph%20x%5E3.gif
thats the x^3 and you have to add +1 to the y

Answer 3

the x-1 is a line and you have to subtract 1

Answer 4

or you might be able to just get away with shifting it all tothe right by 1 :)

Answer 5

huh :S?

Answer 6

i know how the graphs look i just dont know where they touch and...which is upper and which is lower

Answer 7

for the intersection points just put (x-1) in y

Answer 8

It is possibly an error in writing down the question or an error by your instructor: There is no defined error. It goes to infinity.

Answer 9

no defined area, I meant to say.

Answer 10

is the x=-1 and x=1 points of intersection?

Answer 11

OK, I got it the x's are considered boundaries.

Answer 12

oh wow thanks amistre :)

Answer 13

i drew the same thing...

Answer 14

in the boundet area of x=-1 to x =1 there are no intersection points

Answer 15

i dont think the part beneath the xaxis hurts you; but you might wanna shift all this to the right by 1

Answer 16

Don't bother shifting Amistre, it works on in the wash.

Answer 17

y=(x-1)^3+1
y=(x-1)-1

y=x^3 -3x^2 +3x -2
y=x-2

Answer 18

:) i hope so; i can never remember what the outcomes are with left and right of the y axis

Answer 19

\[\int\limits_{-1}^{1}[(x ^{3}+1)-(x -1)]\]

Answer 20

how do you know which is upper and lower?

Answer 21

You can graph it. If you cant use a calculator, you should know the x^3 starts at 1 and goes up and the x starts at -1. So x^3 is on top.

Answer 22

what on top okay...

Answer 23

y=x^3 -3x^2 +3x
y=x-2

{S} (x^3 -3x^2 +3x - (x-2) dx ; [0,2]

{S} (x^3 -3x^2 +2x +2) dx ; [0,2]

x^4/4 -x^3 +x^2 +2x ; at x=0 is 0

(2)^4/4 -(2)^3 +(2)^2 +2(2)

(2)^4/4 -8 +4 +4

2.2.2.2
------ = 4 if i did it right lol
2.2

Answer 24

i got 4 too

Answer 25

andit 4 without the shift ;)

Answer 26

\[\int\limits_{-1}^{1}(x^3+1)-(x-1) dx\]

Answer 27

if i move my graph like puzzle pieces I can make 4 boxes too lol

Answer 28

i did it using that way though

Answer 29

purp; now do it with the functions reversed

{S} (x-1) - (x^3+1) dx

{S} x-1 -x^3-1 dx

{S} x -x^3-2 dx

x^2/2 -x^4/4 -2x

Answer 30

lol thanks so much amistre, chaguanas and nistal! :)

Answer 31

it should be the same answer only negative

Answer 32

oh ok cool...

Answer 33

so dont worry so much about which functions on top or bottom just take the |results| :)

Answer 34

lol

Answer 35

ok thats if we dont have a question sepcifically on area under below x axis or area above x-axis lol