Question

Evaluate the following using properties of integration

$\int_{0}^{4} ( |x - 1| + |x - 2| ) dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int_{0}^{4} ( |x - 1| + |x - 2| ) dx$

Let assume that

$\rm \:f(x) = |x - 1| + |x - 2|$

Let first define the function by using definition of Modulus function.

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - (x - 1) - (x - 2), \: \: \:0 \leqslant x \leqslant 1 } \\ &\sf{ \: \: \: (x - 1) - (x - 2), \: \: \: 1 \leqslant x \leqslant 2}\\ &\sf{ \: \: \: (x - 1) + (x - 2), \: \: \: 2 \leqslant x \leqslant 4} \end{cases}\end{gathered}\end{gathered}$

can be further simplified as

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - 2x + 3, \: \: \:0 \leqslant x \leqslant 1 } \\ &\sf{ \: \: \: \: \: \: \: \: 1, \: \: \: \: \: \: 1 \leqslant x \leqslant 2}\\ &\sf{ \: \: \: 2x - 3, \: \: \: 2 \leqslant x \leqslant 4} \end{cases}\end{gathered}\end{gathered}$

Now, Consider

$\rm \: \displaystyle\int_{0}^{4} f(x) dx$

$\rm \: = \: \displaystyle\int_{0}^{1} f(x) dx + \displaystyle\int_{1}^{2} f(x) dx + \displaystyle\int_{2}^{4} f(x) dx$

$\rm \: = \: \displaystyle\int_{0}^{1} ( - 2x + 3) dx + \displaystyle\int_{1}^{2} 1 dx + \displaystyle\int_{2}^{4} (2x - 3) dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int\rm {x}^{n} \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this result we get

$\rm \: = \: \bigg( - {x}^{2} + 3x \bigg)_0^1 + \bigg(x\bigg)_1^2 + \bigg( {x}^{2} - 3x\bigg)_2^4$

$\rm \: = \: \bigg( - 1 + 3\bigg) + \bigg(2 - 1 \bigg) + \bigg((16 - 12) - (4 - 6) \bigg)$

$\rm \: = \: 2 + 1+ 6$

$\rm \: = \: 9$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\int_{0}^{4} ( |x - 1| + |x - 2| ) dx = 9 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b} f(x) dx \: = \: - \: \displaystyle\int_{b}^{a} f(x) dx}}$

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b} f(x) dx \: = \: \displaystyle\int_{a}^{b} f(y) dy}}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{a} f(x) dx \: = \: \displaystyle\int_{0}^{a} f(a - x) dx}}$

$\boxed{ \rm{ \:\displaystyle\int_{a}^{b} f(x) dx \: = \: \displaystyle\int_{a}^{b} f(a + b - x) dx}}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a} f(x) dx \: = 0, \: \: \rm \: if \: f( - x) = - f(x)}}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^{a} f(x) dx \: = 2\displaystyle\int_{0}^{a} f(x) dx, \: \: \rm \: if \: f(x) = f(x)}}$