# Calculus

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i need help finding area under curve of:2y=sqrt(3x),y=4, and 2y+1x=4

• Calculus -

Did you make a sketch?
Your intersectio points are not "nice" making for messy arithmetic.
the curve and y=4 intersect at (64/3 , 4)
the curve and the line x+2y=4 intersect at appr (1.73, 1.14) and the line x+2y=4 intersects y = 4 at ((-4,4)

So you have to find it in two parts:
Left section:
Area = [integral] (4 - (-1/2)x + 2) ) dx from -4 to 1.73
= (4x + (1/4)x^2 - 2x ) dx from -4 to 1.73
= you do the arithmetic

Right section:
Area = Area = [integral] (4 - (1/2)(3x)^(1/2) ) dx from 1.73 to 64/3
= (4x - (1/9)(3x)^(3/2) ) from 1.73 to 64/3
= you do the arithmetic

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