Find the area of the region bound by the curves 2y= 4√x , y=3 & 2y+1x=5

Answer 1

Area s = ∫₂³(y²/4 - 5 + 2y)dy = (y³/12 - 5y + y²)₂³ = (27/12 -15 + 9) - (8/12 - 19/12 - 10 + 4) = 19/12.

Answer 2

the 3 equations as functions of x are:

y=3,y=2sqrx and y =-.5x +2.5

enclose an area between a flat horizontal line, the upward half of a rightward opening parabola, and a downward sloping line

graph them with a graphing calculator

and it shows two areas enclosed by those 3 lines

one above the other, so two possible answers, altho 2nd treats y=3 as superfluous

for what it's worth several people have come up with different answers

19/12, 23/60=about .26

another about .38

majority favors <1, minority >1

and that's just for the top area

not that math is democratic,

visually from the graph, the top area definitely appears less than 1/2, definitely not>1

Answer 3

Intersection points

2y = 4√x and y = 3 intersect at √x=3/2 which means that x = 9/4

The intersection point is (9/4, 3)

2y + x = 5 and y = 3 intersect at x = –1

The intersection point is (–1, 3)

2y = 4√x and 2y + x = 5 intersect at 4√x = 5 – x

which means that x + 4√x – 5 = 0

leading to (√x + 5)(√x – 1) = 0 and x = 1

The intersection point is (1, 2)

Boundaries, Integrals and Area

Our 3 boundaries for the enclosed area come from the functions

y = 3, y = 2√x and y = ½(5 – x)

On the interval [–1, 1]  3 ≥ ½(5 – x) ≥ 2√x

On the interval [1, 9/4]  3 ≥ 2√x ≥ ½(5 – x)

So the required area is found by solving the integrals

Desmos graph: desmos.com/calculator/mil4iymhbd

However, rewriting the equations as functions of y:

2y = 4√x becomes x= y²/4 and 2y + x = 5 becomes x = 5 – 2y

then integrating between these curves with respect to y

is a more elegant approach leading to a simpler integral and area calculation.