anonymous
  • anonymous
Calculate the area bounded:
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
1 Attachment
freckles
  • freckles
A little drawing always helps to setup the integral.
freckles
  • freckles
\[\int\limits_c^d (\text{ \right relation } - \text{ \left relation } ) dy\]

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anonymous
  • anonymous
I made the drawing. Cant identify which are is bounded thoough
anonymous
  • anonymous
|dw:1441747573378:dw|
freckles
  • freckles
where is the other part of your parabola
anonymous
  • anonymous
Dont you square the "x" to make it y=\[\sqrt{x}\]
freckles
  • freckles
|dw:1441747647427:dw|
freckles
  • freckles
|dw:1441747703759:dw|
freckles
  • 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
  • freckles
in other words y^2=x is a parabola
freckles
  • 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
  • anonymous
Aaaaah, I see. I missed the other half.
anonymous
  • anonymous
Am in integrating in respect to Y?
freckles
  • freckles
yep because we are looking at right and left function
Jhannybean
  • 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
  • anonymous
So limits will be 2 and -1?
Jhannybean
  • Jhannybean
Therefore in that respect, \(y^2 = x \iff y = \pm \sqrt{x}\) haha. sorry if that interrupted you.....
anonymous
  • anonymous
Lol ,Jhannybean. You're good.
freckles
  • freckles
yes the limits are y=-1 to y=2
anonymous
  • anonymous
\[\int\limits_{-1}^{2} \sqrt{y}-(y+5) dy\]
freckles
  • freckles
why sqrt(y)?
freckles
  • freckles
it it x=y^2 not x=sqrt(y)
anonymous
  • anonymous
oopsyes i meant y^2
freckles
  • freckles
\[\int\limits_{-1}^{2} (y^2-(y-5)) dy\]
anonymous
  • anonymous
\[∫y ^{2}−(y+5)dy\]
freckles
  • freckles
also you have x=y-5 not x=y+5
anonymous
  • anonymous
When you flip it turns to a minus?
freckles
  • freckles
huhh
freckles
  • freckles
y=x+5 subtracting 5 on both sides gives y-5=x
freckles
  • 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
  • Jhannybean
You're trying to keep everything consistent. If you're solving for x, then solve all your intersecting functions for x only.
Jhannybean
  • Jhannybean
\[\int_a^b A(y)dy\]
anonymous
  • anonymous
\[\int\limits_{-1}^{2} \frac{ y ^{3} }{ 3 } - \frac{ y ^{2} }{ 2 }-5y\]
freckles
  • freckles
too many signs and you are suppose to drop the integral sign after integrating
freckles
  • freckles
\[\int\limits\limits_{-1}^{2} (y^2-(y-5)) dy=[\frac{y^3}{3}-\frac{y^2}{2}+5y]_{-1}^{2}\]
anonymous
  • anonymous
shouldn't that 5y be -5y?
freckles
  • freckles
no -(-5) =5 not -5
anonymous
  • anonymous
oooooooh
Jhannybean
  • Jhannybean
\[\int\limits_{-1}^{2} (y^2-(y-5)) dy = \int_{-1}^2 (y^2-y+5)dy\]
Jhannybean
  • Jhannybean
Remember when you're integrating to simplify EVERYTHING inside your ( ) first before integrating. It makes integrating a lot easier and error free.
anonymous
  • anonymous
Got it.
anonymous
  • anonymous
I got a final answer of (32/3)-(35/6)=29/6
freckles
  • freckles
33/2 is what I have but I could have made an error
freckles
  • 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
  • freckles
I think I see what you messed on
anonymous
  • 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
  • Jhannybean
Good job.

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