Calculus 2 Area of a Region

Question

Can someone walk me through this? I'm not understanding.

 

Consider the region enclosed by the graphs of 2y=5√x , y=8 , 2y+1x=6

The area A of this region could be computed in two ways:

1) By integrating with respect to x, A = ∫ from a to b of h₁(x)dx + ∫ from b to c of h₂(x)dx

 

a= ?

b= ?

c= ?

h_{1(x)= ?}

h₂(x)= ?

The area A?

 

2) By integrating with respect to y, A = ∫ from d to e of h₃(y)dy

 

d= ?

e= ?

h₃(y)= ?

A= ?

Answer 1

The three functions can be written as

y = 2.5√x

y = 8

y = 3 − x/2

Integration with respect to x

Intersection of y = 8 & y = 3 − x/2 comes at (-10, 8).

Intersection of y = 2.5√x & y = 3 − x/2 comes at (1, 2.5).

Intersection of y = 2.5√x & y = 8 comes at (10.24, 8).

Next, construct ∫_{(-10 to 1)} [8 − (3 − x/2)]dx + ∫_{(1 to 10.24)}[8 − 2.5x^0.5]dx.

Translate to [5x + x²/4|_{(-10 to 1)}] + [8x− (5/3)x^3/2|_{(1 to 10.24)}], which

gives Area as 30.25 + 20.973333333 or 51.223333333.

Integration with respect to y

The three functions can also be written as

x = y²/6.25

y = 8

x = 6 −2y

Then construct ∫_{(2.5 to 8)} [1 − 6 + 2y]dy + ∫_{(2.5 to 8)}[y²/6.25 −1]dy.

Simplify to ∫_{(2.5 to 8)}[y²/6.25 + 2y −6]dy which integrates to

[y³/18.75 + y² − 6y| _{(2.5 to 8)}] equal to 51.223333333.

Then obtain:

a = -10

b = 1

c = 10.24

h₁(x) = 5 + x/2

h₂(x) = 8 − 2.5x^1/2

A = 51.223333333

d = 2.5

e = 8

h₃(y) = y²/6.25 + 2y −6

A = 51.223333333