Question

Express \(D\) as a union of regions of type I (or) type II and evaluate the integral .

Short answer

The process of finding a function \(g\left( x \right)\) the derivative of which, \(Dg\left( x \right)\), is equal to a given function \(f\left( x \right)\).

Step-by-step solution

Graphical Representation.

Consider the integral,

The region of the integration can be partitioned into four region of type 1,

i.e., \(D = {D_1} \cup {D_2} \cup {D_3} \cup {D_4}\)

Name the original integral \(I\) with this partition of \(D\). \(I\) cab be rewritten as the sum of four integrals each with simpler boundaries.

Sub Region.

\({D_1} = \left\{ {y = 1,x = - 1,y = x + 1} \right\}\)

\({D_2} = \left\{ {x = - 1,y = - 1,y = - x - 1} \right\}\)

\({D_3} = \left\{ {y = - 1,x = 1,y = x - 1} \right\}\)

\({D_4} = \left\{ {x = 1,y = 1,y = - x + 1} \right\}\)

Therefore,

\(I = \int\limits_{ - 1}^0 {{x^2}} \int\limits_{x + 1}^1 {dydx} + \int\limits_{ - 1}^0 {{x^2}} \int\limits_{ - 1}^{ - x - 1} {dydx} + \int\limits_0^1 {{x^2}} \int\limits_{ - 1}^{x - 1} {dydx} + \int\limits_0^1 {{x^2}} \int\limits_{ - x + 1}^1 {dydx} \)

Integration.

\(\begin{array}{c}I = \int\limits_{ - 1}^0 {{x^2}} \left( {1 - x - 1} \right)dx + \int\limits_{ - 1}^0 {{x^2}} \left( { - x - 1 + 1} \right)dx + \int\limits_0^1 {{x^2}} \left( {x - 1 + 1} \right)dx + \int\limits_0^1 {{x^2}} \left( {1 + x - 1} \right)dx\\ = \int\limits_{ - 1}^0 { - {x^3}} dx + \int\limits_{ - 1}^0 { - {x^3}} dx + \int\limits_0^1 {{x^3}} dx + \int\limits_0^1 {{x^3}} dx\\ = 2\left[ {\int\limits_{ - 1}^0 { - {x^3}} dx + \int\limits_0^1 {{x^3}} dx} \right]\end{array}\)

Integrate on \(x\),

\(\begin{array}{l}I = 2\left[ { - \frac{1}{4}\left[ {{x^4}} \right]_{ - 1}^0 + \frac{1}{4}\left[ {{x^4}} \right]_0^1} \right]\\I = 2\left[ {0 + \frac{1}{4} + \frac{1}{4} - 0} \right]\\I = 1\end{array}\)

Therefore, the value of integral is \(1\).