Question
Evaluate: $\int\limits_{0}^{4}(|\text{x}|+|\text{x} - 2| + |\text{x} - 4 |)\text{dx}.$

Answer

Let $\text{I} =\int_{0}^{4}(\text{x}| + |\text{x} - 2 | + |\text{x} - 4 |)\text{dx|}$
$ = \int_{0}^{4}|\text{x}|\text{dx} + \int_{0}^{4}|\text{x} - 2|\text{dx} + \int_{0}^{4}|\text{x} - 4|\text{dx}$
$ =\int_{0}^{4}|\text{x}|\text{dx} + +\bigg[\int_{0}^{2}|\text{x} - 2|\text{dx} + \int_{2}^{4}|\text{x} - 2|\text{dx}\bigg] + \int_{0}^{4}|\text{x} - 4 |\text{dx}$ [By properties]
$ =\int_{0}^{4}\text{x dx} + \int_{0}^{2} - (\text{x} - 2 ) \text{dx} + \int_{2}^{4}(\text{x} - 2 )\text{dx} + \int_{0}^{4} - (\text{x} - 4)\text{dx}$
$ \begin{bmatrix}\because|\text{x}| =\text{x},\text{if }0\leq\text{x}\leq4 \\[0.3em] |\text{x - 2}| = -(\text{x} - 2),\text{if }0\leq\text{x}\leq2 \\[0.3em] |\text{x} - 2 | = (\text{x} - 2 ),\text{ if }2\leq\text{x}\leq4 & \\|\text{x} - 4| = -(\text{x} - 4),\text{ if }0\leq\text{x} \leq4 \end{bmatrix}$
$ =\bigg[\frac{\text{x}^{2}}{2}\bigg]^{4}_{0} - \bigg[\frac{(\text{x} - 2)^{2}}{2}\bigg]_{0}^{2} + \bigg[\frac{(\text{x} - 2 )^{2}}{2}\bigg]_{2}^{4} - \bigg[\frac{(\text{x} - 4)^{2}}{2}\bigg]_{0}^{4}$
$ = \frac{1}{2}\times16-\frac{1}{2}\times(0 - 4) + \frac{1}{2}(4- 0)-\frac{1}{2}\times(0- 16 )$
= 8 + 2 + 2 + 8 = 20.

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