How do you find the area between the curves $x+3y=21$ and $x+7=y^2$ ?

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Answer 1 · 2015-03-29T00:14:49

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Solve each function for $x$

$x=21-3y$

$x=y^2-7$

Now find out where the two curves intersect.

$y^2-7=21-3y$

$y^2+3y-28=0$

$(y+7)(y-4)=0$

$y=-7$ and $y=4$

The integral for the area is

$int_-7^4 21-3y-(y^2-7)dy$

$int_-7^4 21-3y-y^2+7dy$

$int_-7^4 28-3y-y^2dy$

Integrating we get

$28y-3/2y^2-1/3y^3$

Now evaluate

$[28(4)-3/2(4)^2-1/3(4)^3]$
$-[28(-7)-3/2(-7)^2-2/3(-7)^3]$

$[112-24-64/3]-[-196-147/2+343/3]$

$[200/3]-[-931/6] =200/3+931/6=1331/6~=221.83$

How do you find the area between the curves x+3y=21x+3y=21 and x+7=y^2x+7=y^2?