Math Help!?

Question

Find the area of the region bounded by the graph of y^3 = x^2 and the chord joining the points (-1,1) and (8,4)

Answer 1

y =
= (x)^(2/3)
Area = ∫y dx = ∫(x)^(2/3) dx = [3/5 x ^ 5/3]
taking the boundaries x = -1,x=8

A1 = 3/5(32+1)= 99/5 = 19.8 (Area between curve and x axis)

Area between chord and x axis =>
eqn of chord => y = x/3 +4/3
A2 = ∫y dx = [x^2/6 + (4/3)x]
take the boundaries x=-1,x=8 A2 = 135/6 (Area between chord and x axis)

Area = A2 - A1 => 45/2-99/5 = (225-198)/10 = 27/10 = 2.7

Answer = 2.7

Answer 2

First off, y^3 = x^2 is the same as
y = x^(2/3). For teaching purposes I'm going to rename this
f(x) = x^(2/3).

We require the function (which we will call g(x)) joining the points (-1, 1) and (8, 4) . This is a high school problem, so I'll skip the details and tell you that the function which represents this is

g(x) = (1/3)x + (4/3)

Normally we would equate f(x) to g(x) to determine the bounds of integration, but it's already given that it is a chord, which means we take the x-values of the chord. Our bounds of integration should be -1 to 8.

Now, we determine which of f(x) and g(x) is greater, between -1 and 8. The reason is because our area formula is going to be

A = Integral (-1 to 8, [higher function] - [lower function] dx )

All we have to do is test one value in between -1 and 8; test
x = 0. Then
f(0) = 0^(2/3) = 0
g(0) = (1/3)(0) + (4/3) = 4/3

So clearly, g(x) is greater on the interval. Our area formula is then

A = Integral (-1 to 8, [g(x) - f(x)] dx )

Plug in our functions,

A = Integral (-1 to 8, { (1/3)x + (4/3) - x^(2/3) } dx )

And now we use the reverse power rule.

A = [ (1/3)(1/2)x^2 + (4/3)x - (3/5)x^(5/3) ] {evaluated from -1 to 8}

Let's simplify that, for easier calculation.

A = [ (1/6)x^2 + (4/3)x - (3/5)x^(5/3) ] {from -1 to 8}

A = [ (1/6)(8)^2 + (4/3)(8) - (3/5)(8)^(5/3) ] -
[ (1/6)(-1)^2 + (4/3)(-1) - (3/5)(-1)^(5/3) ]

A = [ (1/6)(64) + (32/3) - (3/5)(32) ] - [(1/6) - (4/3) - (3/5)]

A = [(64/6) + (32/3) - (96/5)] - [(1/6) - (4/3) - (3/5)]

Notice how, in each set of brackets, the first fractions are over 6, the second fractions are over 3, and the third fractions are over 5. Notice also that we have subtraction; this means we can subtract each fraction component-wise.

A = (64/6 - 1/6) + (32/3 - (-4/3)) + (-96/5 + 3/5)
A = (63/6) + (36/3) + (-93/5)
A = (21/2) + (12) - (93/5)
A = (105/10) + (120/10) - (186/10)
A = 220/10 + 186/10
A = 406/10 = 203/5