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Evaluate: ∫01x2-2x2+1.dx - Mathematics and Statistics

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प्रश्न

Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`

योग

उत्तर

`int_0^1 (x^2 - 2)/(x^2 + 1).dx`

= `int_0^1 ((x^2 + 1) - 3)/(x^2 + 1).dx`

= `int_0^1(1 - 3/(x^2 + 1)).dx`

= `int_0^1 1.dx - int_0^1 3/(x^2 + 1).dx`

= `[x]_0^1 - [3 tan^-1x]_0^1`

= (1 – 0) – (3 tan–11 – 3 tan–10)

= `1 - 3(pi/4) - 0`

= `1 - (3pi)/(4)`.

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Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Exercise 4.2 [पृष्ठ १७१]

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