Find the area of the region bounded by the graphs of the given curves

Question

y=x^3/2 and y=(x^2)/32

Answer 1

First, let's find the left and right limits of integration, i.e., the x values for which the two graphs touch.

We have to solve x^3/2 = x²/32.

Let's subtract x^3/2 from both sides of the equation, to make one side be zero

(and then I'll write the exponent 3/2 as 1.5):

0 = x²/32 - x^3/2 = x²/32 - x^1.5

Then factor the right side into two factors:

(and then I'll write the exponent.5 as a square root sign √)

0 = x²/32 - x^1.5 = x^1.5((x^.5)/32 - 1) = x^1.5((√x)/32 - 1)

The two factors are x^1.5 and (√x)/32 - 1.

The product of the two factors will be zero if we can make either one of the factors zero.

So we have two solutions:

the solution x=0 makes the first factor (x^1.5) be zero; and

the solution x = 1024 = 32² makes the second factor ((√x)/32 - 1) be zero, because √1024 = 32

So our left and right limits of integration are x=0 and x=1024.

Let's find out which curve is higher than the other in the interval from x=0 to x=1024.

We can pick any x in this interval, but it will make the calculations simpler if we pick an x that is a perfect square such as x=4 (or x=9 or x=16, etc.) because we will have to evaluate x^1.5 = x(√x)

So at x=4,

x^1.5 = 4^1.5 = 4(√4) = 4 times 2 = 6

x²/32 = 4²/32 = 16/32 = 1/2

So at x=4, the curve x^1.5 is higher than the curve x²/32.

But that means that the curve x^1.5 is higher than the curve x²/32 throughout the entire interval from x=0 to x=1024, because the two graphs do not cross each other at any point between x=0 and x=1024.

The area between the two curves is ∫ (from x=0 to x=1024) higher curve - lower curve dx

= ∫ (from x=0 to x=1024) x^1.5 - x²/32 dx

Integrating this integral, we have

x^2.5/2.5 - x³/(3 times 32) = x^2.5/2.5 - x³/96.

Plugging in the two limits of integration (thank goodness the left one is zero!), the area between the two curves is

(1024^2.5/2.5 - 1024³/96) minus (0^2.5/2.5 - 0³/96)

= (1024^2.5/2.5 - 1024³/96) minus 0

= 1024^2.5/2.5 - 1024³/96

= (1024 times 1024 times 32)/2.5 - 1024³/96

= 33554432/2.5 - 1024³/96

= 13421772.8 - (11184810 + 2/3)

= 2236962 + 2/15

Answer 2

x^1.5 = (x^2)/32

0^1.5 = 0 = (0^2)/32

(1024)^1.5=32768 = (1024^2)/32

x=0,1024 = integration limits

integrate (x^1.5 - (x^2)/32)

=(x^(1.5+1))/(1.5+1) - (x^(2+1))/32(2+1)

= (x^2.5)/2.5 - (x^3)/96

evaluate from 0 to 1024

(1024)^2.5/2.5 - (1024)^3/96

= 13,421,772.8- 11,184,810.67

=2,236,962.13

=about 2 1/4 million units^2