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anonymous
 one year ago
Calculate the area bounded:
anonymous
 one year ago
Calculate the area bounded:

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2A little drawing always helps to setup the integral.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_c^d (\text{ \right relation }  \text{ \left relation } ) dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I made the drawing. Cant identify which are is bounded thoough

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441747573378:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2where is the other part of your parabola

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dont you square the "x" to make it y=\[\sqrt{x}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441747647427:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441747703759:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2no 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
 one year ago
Best ResponseYou've already chosen the best response.2in other words y^2=x is a parabola

freckles
 one year ago
Best ResponseYou've already chosen the best response.2anyways you have the right relation is given by x=y^2 and the left relation is given by x=y5 and I think it should be clear what the limits are

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aaaaah, I see. I missed the other half.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am in integrating in respect to Y?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep because we are looking at right and left function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Think 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
 one year ago
Best ResponseYou've already chosen the best response.0So limits will be 2 and 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Therefore in that respect, \(y^2 = x \iff y = \pm \sqrt{x}\) haha. sorry if that interrupted you.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol ,Jhannybean. You're good.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yes the limits are y=1 to y=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{2} \sqrt{y}(y+5) dy\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it it x=y^2 not x=sqrt(y)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{1}^{2} (y^2(y5)) dy\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2also you have x=y5 not x=y+5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you flip it turns to a minus?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2y=x+5 subtracting 5 on both sides gives y5=x

freckles
 one year ago
Best ResponseYou've already chosen the best response.2x=y5 is another way to say y=x+5 but we solved for x because we are looking at our functions in terms of y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're trying to keep everything consistent. If you're solving for x, then solve all your intersecting functions for x only.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{2} \frac{ y ^{3} }{ 3 }  \frac{ y ^{2} }{ 2 }5y\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2too many signs and you are suppose to drop the integral sign after integrating

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits\limits_{1}^{2} (y^2(y5)) dy=[\frac{y^3}{3}\frac{y^2}{2}+5y]_{1}^{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0shouldn't that 5y be 5y?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{2} (y^2(y5)) dy = \int_{1}^2 (y^2y+5)dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember when you're integrating to simplify EVERYTHING inside your ( ) first before integrating. It makes integrating a lot easier and error free.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got a final answer of (32/3)(35/6)=29/6

freckles
 one year ago
Best ResponseYou've already chosen the best response.233/2 is what I have but I could have made an error

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits\limits\limits_{1}^{2} (y^2(y5)) dy=[\frac{y^3}{3}\frac{y^2}{2}+5y]_{1}^{2} \\ =[\frac{2^3}{3}\frac{2^2}{2}+5(2)][\frac{(1)^3}{3}\frac{(1)^2}{2}+5(1)] \\ =[\frac{8}{3}2+10][\frac{1}{3}\frac{1}{2}5] \\ =\frac{8}{3}+\frac{1}{3}+\frac{1}{2}2+10+5 \\ =\frac{9}{3}+\frac{1}{2}+13 \\ =3+13+\frac{1}{2} \\ =16+\frac{1}{2} \\ =\frac{32}{2}+\frac{1}{2} \\ =\frac{33}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I think I see what you messed on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see what i did wrong. I got my signs mixed up on the right side. But yes its 33/2 thank you both!
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