Calculate the area bounded:

- anonymous

Calculate the area bounded:

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- anonymous

##### 1 Attachment

- freckles

A little drawing always helps to setup the integral.

- freckles

\[\int\limits_c^d (\text{ \right relation } - \text{ \left relation } ) dy\]

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## More answers

- anonymous

I made the drawing. Cant identify which are is bounded thoough

- anonymous

|dw:1441747573378:dw|

- freckles

where is the other part of your parabola

- anonymous

Dont you square the "x" to make it y=\[\sqrt{x}\]

- freckles

|dw:1441747647427:dw|

- freckles

|dw:1441747703759:dw|

- freckles

no y^2=x doesn't mean just y=sqrt(x)
\[x=y^2 \text{ is } \sqrt{x}=y \text{ if } y>0 \\ x=y^2 \text{ is } -\sqrt{x}=y \text{ if } y<0\]

- freckles

in other words y^2=x is a parabola

- freckles

anyways you have the right relation is given by x=y^2
and the left relation is given by x=y-5
and I think it should be clear what the limits are

- anonymous

Aaaaah, I see. I missed the other half.

- anonymous

Am in integrating in respect to Y?

- freckles

yep because we are looking at right and left function

- Jhannybean

Think about it, if we turned you sideways and half your body was missing and someone asked, "what would he look like if he was turned right side up?" and we responded "only what you can see", they wouldnt think of you as a complete person would they? xD

- anonymous

So limits will be 2 and -1?

- Jhannybean

Therefore in that respect, \(y^2 = x \iff y = \pm \sqrt{x}\) haha. sorry if that interrupted you.....

- anonymous

Lol ,Jhannybean. You're good.

- freckles

yes the limits are y=-1 to y=2

- anonymous

\[\int\limits_{-1}^{2} \sqrt{y}-(y+5) dy\]

- freckles

why sqrt(y)?

- freckles

it it x=y^2 not x=sqrt(y)

- anonymous

oopsyes i meant y^2

- freckles

\[\int\limits_{-1}^{2} (y^2-(y-5)) dy\]

- anonymous

\[∫y ^{2}−(y+5)dy\]

- freckles

also you have x=y-5 not x=y+5

- anonymous

When you flip it turns to a minus?

- freckles

huhh

- freckles

y=x+5
subtracting 5 on both sides gives
y-5=x

- freckles

x=y-5 is another way to say y=x+5
but we solved for x because we are looking at our functions in terms of y

- Jhannybean

You're trying to keep everything consistent. If you're solving for x, then solve all your intersecting functions for x only.

- Jhannybean

\[\int_a^b A(y)dy\]

- anonymous

\[\int\limits_{-1}^{2} \frac{ y ^{3} }{ 3 } - \frac{ y ^{2} }{ 2 }-5y\]

- freckles

too many signs
and you are suppose to drop the integral sign after integrating

- freckles

\[\int\limits\limits_{-1}^{2} (y^2-(y-5)) dy=[\frac{y^3}{3}-\frac{y^2}{2}+5y]_{-1}^{2}\]

- anonymous

shouldn't that 5y be -5y?

- freckles

no -(-5) =5 not -5

- anonymous

oooooooh

- Jhannybean

\[\int\limits_{-1}^{2} (y^2-(y-5)) dy = \int_{-1}^2 (y^2-y+5)dy\]

- Jhannybean

Remember when you're integrating to simplify EVERYTHING inside your ( ) first before integrating. It makes integrating a lot easier and error free.

- anonymous

Got it.

- anonymous

I got a final answer of (32/3)-(35/6)=29/6

- freckles

33/2 is what I have
but I could have made an error

- freckles

\[\int\limits\limits\limits_{-1}^{2} (y^2-(y-5)) dy=[\frac{y^3}{3}-\frac{y^2}{2}+5y]_{-1}^{2} \\ =[\frac{2^3}{3}-\frac{2^2}{2}+5(2)]-[\frac{(-1)^3}{3}-\frac{(-1)^2}{2}+5(-1)] \\ =[\frac{8}{3}-2+10]-[\frac{-1}{3}-\frac{1}{2}-5] \\ =\frac{8}{3}+\frac{1}{3}+\frac{1}{2}-2+10+5 \\ =\frac{9}{3}+\frac{1}{2}+13 \\ =3+13+\frac{1}{2} \\ =16+\frac{1}{2} \\ =\frac{32}{2}+\frac{1}{2} \\ =\frac{33}{2}\]

- freckles

I think I see what you messed on

- anonymous

I see what i did wrong. I got my signs mixed up on the right side. But yes its 33/2 thank you both!

- Jhannybean

Good job.

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