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October 22, 2014

October 22, 2014

Posted by **Chrissy** on Thursday, November 21, 2013 at 10:16pm.

2y=4*sqrt(x) , y=5 and 2y+4x=8

please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't arrive at the right area between the curves. Thanks in advance!

- Calculus-Area between curves -
**Reiny**, Thursday, November 21, 2013 at 10:41pmfirst of all reduce the equation to simplest form:

y = 2√x

y = 5

y + 2x = 4

Your sketch should look like a "triangle" with one side slightly curved

see:

http://www.wolframalpha.com/input/?i=plot+y%3D2*sqrt%28x%29+%2C+y%3D5+%2C+y%2B2x%3D4

we need all the intersection points,

I assume you know how to do that, ....

but I will do the hardest one:

intersect y = 2√x and y + 2x = 4

then 2√x = (4-2x)

or

√x = 2-x

by inspection, x = 1 , then in the originals , y = 2

point of intersection is (1,2)

Hoping that you get (-1/2, 5) and ( 25/4,5) for the other points

so area

= ∫(5 - (4-2x)) dx from -1/2 to 1 + ∫(5 - 2x^(1/2) dx from 1 to 25/4

= (x + x^2)| from -1/2 to 1 + (5x - 4x^(-1/2) )| from 1 to 25/4

= 1 + 1 - (-1/2 + 1/4) + (125/4 - 8/5 - (5-4) )

= 2 + 1/4 + 125/4 - 8/5 - 1

= 309/10

you better check my arithmetic on that one, should have written it out on paper first.

- Calculus-Area between curves -
**Chrissy**, Thursday, November 21, 2013 at 11:15pmThanks I figured it out my answer was 9, and it was correct

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