Assuming the area under the curve y=x^2 goes from 0 to n.
Area = INT[x^2] dx from 0 to n = 100
(1/3)*x^3 from 0 to n = 100
(1/3)*n^3 = 100
n^3 = 300
n = (300)^(1/3)
x_max = n = 6.69...
Also assuming y(1) means y(first_interval), not literally y(1) = 1^2 = 1.
a) y(1) = x^2
= [(1/10) * n]^2
= [(1/10)*(300)^(1/3)]^2
= (1/100)*(300)^(2/3)
Also assuming y(11) means y(11th_interval), not literally y(11) = 11^2 = 121.
The nth interval is the last interval where x is at the far right or maximum(x_max).
b) y(11) = y(n)
= n^2
= [(300)^(1/3)]^2
= (300)^(3/2)
= 6.69...
c) y = x^z
Area = INT[x^z] dx from 0 to n = 100
(1/(z+1)) * x^(z+1) from 0 to n = 100
(1/(z+1)) * n^(z+1) = 100
n^(z+1) = 100 * (z+1)
n = [ 100 * (z+1) ]^(1/(z+1))
x_max = n
y(n) = y(x_max) = (x_max)^2
y(n) = [[ 100 * (z+1) ]^(1/(z+1)) ]^2
y(n) = [ 100 * (z+1) ]^(2/(z+1))
