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Evaluate the definite integrals in Exercises 25 to 33.

Q33. $\int_1^4[|x-1| + |x-2| + |x-3|]dx$

Answers (1)

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Given integral $\int_1^4[|x-1| + |x-2| + |x-3|]dx$

So, we split it in according to intervals they are positive or negative.

$= \int_{1}^4 |x-1| dx + \int_{1}^4 |x-2| dx + \int^4_{1} |x-3| dx$

$= I_{1}+I_{2}+I_{3}$

Now,

$I_{1} = \int^4_{1}|x-1| dx = \int^4_{1} (x-1)dx$

$\because$ as $(x-1)$ is positive in the given x -range $[1,4]$

$=\left [ \frac{x^2}{2}-x\right ]^4_{1} = \left [ \frac{4^2}{2}-4 \right ] - \left [ \frac{1^2}{2}-1 \right ]$

$= \left [ 8-4 \right ] - [-\frac{1}{2}] = 4+\frac{1}{2} = \frac{9}{2}$

Therefore, $I_{1} = \frac{9}{2}$

$I_{2} = \int^4_{1}|x-2| dx = \int^2_{1} (2-x)dx +\int^4_{2} (x-2)dx$

$\because$ as $(x-2)\geq 0$ is in the given x -range $[2,4]$ and $\leq 0$ in the range $[1,2]$

$=\left [ 2x - \frac{x^2}{2}\right ] ^2_{1} + \left [ \frac{x^2}{2} -2x\right ] ^4_{2}$

$= \left \{ \left [ 2(2)-\frac{2^2}{2} \right ] - \left [ 2(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-2(4) \right ] - \left [ \frac{2^2}{2}-2(2) \right ] \right \}$

$= [4-2-2+\frac{1}{2}] +[8-8-2+4]$

$= \frac{1}{2}+2 =\frac{5}{2}$

Therefore, $I_{2} = \frac{5}{2}$

$I_{3} = \int^4_{1}|x-3| dx = \int^3_{1} (3-x)dx +\int^4_{3} (x-3)dx$

$\because$ as $(x-3)\geq 0$ is in the given x -range $[3,4]$ and $\leq 0$ in the range $[1,3]$

$=\left [ 3x - \frac{x^2}{2}\right ] ^3_{1} + \left [ \frac{x^2}{2} -3x\right ] ^4_{3}$

$= \left \{ \left [ 3(3)-\frac{3^2}{2} \right ] - \left [ 3(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-3(4) \right ] - \left [ \frac{3^2}{2}-3(3) \right ] \right \}$

$= [9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$= [6-4]+\frac{1}{2} =\frac{5}{2}$

Therefore, $I_{3} = \frac{5}{2}$

So, We have the sum $= I_{1}+I_{2}+I_{3}$

$I = \frac{9}{2}+\frac{5}{2}+\frac{5}{2} = \frac{19}{2}$

Posted by

Divya Prakash Singh

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