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NCERT Class XII Mathematics Chapter - - Solutions

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Evaluate the definite integrals in Exercises 253 to 261.
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Question : 261 of 271
Marks: +1, -0
∫14[∣x−1∣+∣x−2∣+∣x−3∣]dx\int\limits_{1}^{4} \left[ \left| x-1 \right| + \left| x-2 \right| + \left| x-3 \right| \right] dx
Solution:  
Let I=∫14[∣x−1∣+∣x−2∣+∣x−3∣]dxI = \int\limits_{1}^{4} \left[ \left| x-1 \right| + \left| x-2 \right| + \left| x-3 \right| \right] dx
Define
|x – 1| = x – 1, when x – 1 ≥ 0 i.e., x ≥ 1
|x – 2| = x – 2, when x – 2 ≥ 0 i.e., x ≥ 2
|x – 2| = – (x – 2), when x – 2 ≤ 0 i.e., x ≤ 2
|x – 3| = – (x – 3), when x – 3 ≤ 0 i.e., x ≤ 3
|x – 3| = (x – 3), when x – 3 ≥ 0 i.e., x ≥ 3
⇒I=∫14(x−1) dx−∫12(x−2) dx+∫24(x−2) dx−∫13(x−3) dx+∫34(x−3) dx\Rightarrow I = \int\limits_{1}^{4} (x-1) \, dx - \int\limits_{1}^{2} (x-2) \, dx + \int\limits_{2}^{4} (x-2) \, dx - \int\limits_{1}^{3} (x-3) \, dx + \int\limits_{3}^{4} (x-3) \, dx
=[x22−x]14−[x22−2x]12+[x22−2x]24−[x22−3x]13+[x22−3x]34= \left[ \frac{x^2}{2} - x \right]_{1}^{4} - \left[ \frac{x^2}{2} - 2x \right]_{1}^{2} + \left[ \frac{x^2}{2} - 2x \right]_{2}^{4} - \left[ \frac{x^2}{2} - 3x \right]_{1}^{3} + \left[ \frac{x^2}{2} - 3x \right]_{3}^{4}
=[(162−12)−(4−1)]−[(42−12)−(4−2)]+[(162−42)−(8−4)]−[(92−12)−(9−3)]+[(162−92)−(12−9)]= \left[ \left( \frac{16}{2} - \frac{1}{2} \right) - (4-1) \right] - \left[ \left( \frac{4}{2} - \frac{1}{2} \right) - (4-2) \right] + \left[ \left( \frac{16}{2} - \frac{4}{2} \right) - (8-4) \right] - \left[ \left( \frac{9}{2} - \frac{1}{2} \right) - (9-3) \right] + \left[ \left( \frac{16}{2} - \frac{9}{2} \right) - (12-9) \right]
=[152−32+122−82+72]+[−3+2−4+6−3]= \left[ \frac{15}{2} - \frac{3}{2} + \frac{12}{2} - \frac{8}{2} + \frac{7}{2} \right] + \left[ -3 + 2 - 4 + 6 - 3 \right]
=[232]+[−2]=192= \left[ \frac{23}{2} \right] + \left[ -2 \right] = \frac{19}{2}
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