Question

Will someone please help me with my calculus. Thank you!

Answer 1

\(\color{black}{\displaystyle {\bf c}\aleph\ell {\bf c} \mu \ell \mu \int }\) ?

Answer 2

Ok so I have the equation x=4-y^2 and x=y-2, they want to know the area under the curve, by integrating with respect to x and y

Answer 3

https://www.desmos.com/calculator

Answer 4

maybe they want to know the area bound by these two curves?

Answer 5

the area between the two curves, but I think you are asking the same as what they want

Answer 6

Didn't save, sorry, there.

Answer 7

https://www.desmos.com/calculator/b4wmy6l3ch

Answer 8

The area under the curve isn't the same as the area in between .... but, anyway, they (as I understood) want to know the area between the curves.

Answer 9

yes it says find the area of the region between the two curves

Answer 10

Let's do wrt x first.

Answer 11

ok

Answer 12

Can you find the two y-values at which the functions intersect?

Answer 13

(whether by graphing, or by algebra. I don't really care.)

Answer 14

ok, just a second!

Answer 15

y=-3 and y=2

Answer 16

Yes.

Answer 17

So, if we are integrating wrt x, then your limits of integration are y=-3 and y=2.

Answer 18

You have done area between curves problems before, haven't you?

Answer 19

yes but isn't it goofy with y values?

Answer 20

Oh, I'm doing integration wrt to y, sorry:)

Answer 21

nooooo its ok!

Answer 22

lets keep doing wrt to x

Answer 23

so is the equation just the integral from -3 to 2 (y-2)-(4-y^2)dy?

Answer 24

I think it is the other way around.

Answer 25

The rightmost function is the "upper" (or the "positive" bound.

Answer 26

ohhhh

Answer 27

because were doing with respect to x?

Answer 28

I meant to say that we treat the x-axis as we would treat the y-axis in the normal problem, and this is why I said wrt to x.

(I didn't manage to say what I wanted to say, basically.)

Answer 29

Oh ok

Answer 30

that makes sense

Answer 31

Ok, can you fix the integral for the area of the curve?

Answer 32

(You subtract Right-Left, not the other way around.)

Answer 33

so the integral of -3 to 2 (4-y^2)-(y-2)dy

Answer 34

Awesome!

Answer 35

Can you solve this integral?

Answer 36

(I am pretty sure you can)

Answer 37

yes!!!

Answer 38

sorry I don't have my calculator on me right now so it might just be a second

Answer 39

You can enter it in wolframalpha .... or, by hand.
(It just involves power-rule, not hard.)

Answer 40

Wait?

Answer 41

Sure, take your time.

Answer 42

ok never mind i will do it on the website, I thought we had denominators of 2 when its actually 3 :/

Answer 43

I don't think this is right but -233/6?

Answer 44

not quite.

Answer 45

\(\color{black}{\displaystyle \int _{-3}^{2}[(4-y^2)-(y-2)]dy }\)

\(\color{black}{\displaystyle \int _{-3}^{2}[4-y^2-y+2]dy }\)

\(\color{black}{\displaystyle \int _{-3}^{2}[-y^2-y+6]dy }\)

\(\color{black}{\displaystyle \left.\left(-\frac{y^3}{3}-\frac{y^2}{2}+6y\right)\right|_{y=-3}^{y=2} }\)

Answer 46

Like this.

Answer 47

omg...

Answer 48

What's omg here?

Answer 49

I never changed the order on my paper that is why :D wow I'm silly

Answer 50

Oh, omg in a good way, as of "I get it", not as of "omg, I'm so confused" ...
well, that is an effective result:)

(also, btw wolfram would probably interpret omg as omega)

Answer 51

lol

Answer 52

So, you got the integration wrt y, correct?

Answer 53

Yes, I understand now

Answer 54

No we will do the integration wrt to x.