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Evaluate the definite integral: ∫14[|x-1|+|x-2|+|x-3|]dx - Mathematics

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प्रश्न

Evaluate the definite integral:

`int_1^4 [|x - 1|+ |x - 2| + |x -3|]dx`

योग
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उत्तर

Let `I = int_1^4 (|x - 1| + |x - 2| + |x - 3|)  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 = int_1^4 (x - 1)  dx - int_1^2 (x - 2)  dx + int_2^4 (x - 2)  dx  - int_1^3 (x - 3) dx + int_3^4 (x - 3)  dx`

`= [x^2/2 - x]_1^4 - [x^2/2 - 2x]_1^2 + [x^2/2 - 2x]_2^4 - [x^2/2 - 3x]_1^3 + [x^2/2 - 3x]_3^4`

`= [(16/2 - 1/2) - (4 - 1)] - [(4/2 - 1/2) - (4 - 2)] + [(16/2 - 1/2) - (8 - 4) - [(9/2 - 1/2) - (9 - 3)] + [(16/2 - 9/2) - (12 - 9)]`

`= [15/2 - 3/2 + 12/2 - 8/2 + 7/2] + [-3 + 2 - 4 + 6 - 3]`

`= [23/2] + [-2]`

`= 19/2`

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अध्याय 7: Integrals - Exercise 7.12 [पृष्ठ ३५३]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.12 | Q 33 | पृष्ठ ३५३

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