Question

Evaluate ∫∫xy(x + y)dxdy for x, y ∈ [R] over the region between y = x² and y = x.

Answer 1

The bounded curves are y = x² and y = x.

Common points are given by solving two equations.

x² = x

x(x - 1) = 0

x = 0 or x = 1

if x = 0, y = 0

if x = 1, y = 1

\(\int \int\limits _R xy(x + y)dy\, dx\)

\(= \int \limits_{x = 0}^1 \;\int \limits_{y = x^2}^x xy(x + y)dy\, dx\)

\(= \int \limits_0^1 \int\limits_{x^2}^x x^2y+ xy^2 dy \,dx \)

\(= \int \limits_0^ x x\left [\frac{xy^2}2+\frac{y^3}3\right]_{x^2}^x dx\)

\(= \int \limits _0^1 x \left[\frac{x^3}2 + \frac{x^3}3 - \frac{x^5}2 - \frac{x^6}{3}\right]dx\)

\(= \int \limits_0^1 \frac 56 x^4 - \frac{x^6}2 - \frac{x^7}3 dx\)

\(= \left[\frac 56 \frac{x^5}5 - \frac 12\frac{x^6}7- \frac 13 \frac{x^8}8\right]_0^1\)

\(= \left[\frac 56. \frac{1}5 - \frac 12.\frac{1}7- \frac 13 .\frac{1}8\right]-0\)

\(= \frac 16 - \frac 1{14} - \frac 1{24}\)

\(= \frac 3{56}\)

Answer 2

The bounded curves are y = x² and y = x. The common points are given by solving the two equations.

So, we have

x² = x = x (x – 1) = 0

⇒ x = 0 or 1

when x = 0, we have y = 0 and

when x = 1, y = 1 (from y = x)