Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Evaluate the definite integral $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$
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Answer
Given: $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$
$\Rightarrow \mathrm{I}=\int_{1}^{4}[|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|] \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{1}^{4}[|\mathrm{x}-1|] \mathrm{dx}+\int_{1}^{4}[|\mathrm{x}-2|] \mathrm{d} \mathrm{x}+\int_{1}^{4}[|\mathrm{x}-3|] \mathrm{d} \mathrm{x}$
Let $I = I_1 + I_2 + I_3$
First solve for $I_1$:
$I_{1}=\int_{1}^{4}[|x-1|] d x$
As we can see that $(x-1) \geq 0$ when $1 \leq x \leq 4$
$\Rightarrow \mathrm{I}_{1}=\int_{1}^{4}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{1}=\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{1}^{4}$
$\Rightarrow \mathrm{I}_{1}=\left[\frac{(4)^{2}}{2}-4-\frac{(1)^{2}}{2}+1\right]$
$\Rightarrow \mathrm{I}_{1}=\left[8-4-\frac{1}{2}+1\right]$
$\Rightarrow I_{1}=\left[5-\frac{1}{2}\right]$
$\Rightarrow \mathrm{I}_{1}=\frac{9}{2}$
Now solve for $I_2$:
$I_{2}=\int_{1}^{4}[|x-2|] d x$
As we can see that $(x-2) \leq 0$ when $1 \leq x \leq 2$ and $(x-2) \geq 0$ when $2 \leq x \leq 4$
as, $\left\{\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\right\}$
$\Rightarrow \mathrm{I}_{2}=\int_{1}^{2}-(\mathrm{x}-2) \mathrm{d} \mathrm{x}+\int_{2}^{4}(\mathrm{x}-2) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{\mathrm{x}^{2}}{2}-2 \mathrm{x}\right]_{1}^{2}+\left[\frac{\mathrm{x}^{2}}{2}-2 \mathrm{x}\right]_{2}^{4}$
$\Rightarrow \mathrm{I}_{2}=-\left[\frac{(2)^{2}}{2}-2(2)-\frac{(1)^{2}}{2}+2(1)\right]$ + $\left[\frac{(4)^{2}}{2}-2(4)-\frac{(2)^{2}}{2}+2(2)\right]$
$\Rightarrow \mathrm{I}_{2}=-\left[2-4-\frac{1}{2}+2\right]+[8-8-2+4]$
$\Rightarrow \mathrm{I}_{2}=\left[\frac{1}{2}+2\right]$
$\Rightarrow \mathrm{I}_{2}=\frac{5}{2}$
Now solve for $I_3$:
$I_{3}=\int_{1}^{4}[|x-3|] d x$
As we can see that $(x-3) \leq 0$ when $1 \leq x \leq 3$ and $(x-3) \geq 0$ when $3 \leq x \leq 4$
as, $\left\{\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\right\}$
$\Rightarrow \mathrm{I}_{3}=\int_{1}^{3}-(\mathrm{x}-3) \mathrm{d} \mathrm{x}+\int_{3}^{4}(\mathrm{x}-3) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{1}^{3}+\left[\frac{\mathrm{x}^{2}}{2}-3 \mathrm{x}\right]_{3}^{4}$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{(3)^{2}}{2}-3(3)-\frac{(1)^{2}}{2}+3(1)\right]$ + $\left[\frac{(4)^{2}}{2}-3(4)-\frac{(3)^{2}}{2}+3(3)\right]$
$\Rightarrow \mathrm{I}_{3}=-\left[\frac{9}{2}-9-\frac{1}{2}+3\right]+\left[8-12-\frac{9}{2}+9\right]$
$\Rightarrow I_{3}=\left[2+\frac{1}{2}\right]$
$\Rightarrow I_{3}=\frac{5}{2}$
as $I = I_1 + I_2 + I_3$
$\Rightarrow I=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}$
$\Rightarrow \mathrm{I}=\frac{19}{2}$
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