We want to evaluate
[tex]\int \int (x^{2}+6y)dA[/tex]
over the region bounded by
y = x, y = x³, and x ≥ 0.
The intersection of the curves y=x and y = x³ is determined as
x³ = x
x(x² - 1) = 0
x(x + 1)(x - 1) = 0
x =0, x = -1, x = 1.
Because x ≥ 0, the intersection points are
(0,0) and (1,1)
The region for the integration is shown shaded in the graph shown below.
Let
y₁ = x³, the curve.
y₂ = x, the straight line
Then the integral may be written as
[tex]\int_{0}^{1} dx \int_{y1}^{y2} (x^{2}+6y)dy \\\\ = \int _{0}^{1}dx \int_{x^3} ^{x} (x^{2}+6y)dy \\ [/tex]
The inner integral yields
[tex][x^{2}y+3y^{2}]_{x^{3}}^{x} = x^{3} - x^{5} + 3(x^{2} - x^{6})[/tex]
The outer integral yields
[tex]\int_{0}^{1} (x^{3}-x^{5}+3x^{2}-3x^{6})dx \\\\ = [ \frac{x^{4}}{4} - \frac{x^{6}}{6} +3 (\frac{x^{3}}{3}) -3( \frac{x^{7}}{7})]_{0}^{1} \\\\ = \frac{55}{84} [/tex]
Answer: 55/84 (or 0.655)
