Question

help with a calc problem!

Answer 1

@jim_thompson5910 i ran into another one :/

Answer 2

@freckles

Answer 3

do you know what you are attempting to evaluate?

Answer 4

you've got \[y^2=2x+6~,~ x=y+1\]\[\implies y=\sqrt{2x+6}~,~ y=x-1\]

Answer 5

Now graph these two, find the intersection, integrate

Answer 6

you should graph first to know that you want to find

Answer 7

because the equation depend on it

Answer 8

Yep

Answer 9

so they intersect at (5,4)?

Answer 10

That's one intersection, yeah.

Answer 11

your graph missed some parts

Answer 12

look here
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Answer 13

ohh i see it

Answer 14

so the second one is -2, -2

Answer 15

what do i do next??

Answer 16

@xapproachesinfinity ??

Answer 17

sorry i meant -1,-2

Answer 18

hmm we go this way i guess
we do int from -2 to 4
the equation y=x+1 > y^2=2x+6

Answer 19

@Jhannybean said we integrate??

Answer 20

it also said that i have to have an integral in one variable

Answer 21

@jim_thompson5910

Answer 22

so \[\int_{-2}^4\left [(y-1)-(\frac{y^2-6}{2})\right]dy \]

Answer 23

sorry got disconnected somehow

Answer 24

ohhh okay got it!

Answer 25

hmm have done this in awhile i could be off
but if someone checks this lol :)

Answer 26

@jim_thompson5910 knows this very well(:

Answer 27

yeah :)

Answer 28

i meant better if someone checks this :)
don't why i keep writing nonsense stuff when typing

Answer 29

did you compute the integral

Answer 30

I agree with xapproachesinfinity

\[\large \int_{-2}^4\left [(y-1)-(\frac{y^2-6}{2})\right]dy\]

looks like a good way to do it. You could do it using x and dx, but you'd have to do it in 2 pieces. So this method is more direct.

Answer 31

yeah just about posting the other way with dx

Answer 32

seems to me dy is better

Answer 33

correct

Answer 34

with dx
you would consider an extra equation \[y=-\sqrt{2x+6}\]
since it is part of the area

Answer 35

i would illlustrate the idea by a drawing perhaps