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Evaluate: ∫0113+2x-x2⋅dx - Mathematics and Statistics

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प्रश्न

Evaluate:

`int_0^1 (1)/sqrt(3 + 2x - x^2)*dx`

योग

उत्तर

`int_0^1 (1)/sqrt(3 + 2x - x^2)*dx`

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

= `int_0^1 (1)/sqrt((2)^2 - (x - 1)^2)*dx`

= `[sin^-1 ((x - 1)/2)]_0^1`

= `sin^-1(0) - sin^-1(-1/2)`

= `0 - sin^-1 (-sin  pi/6)`

= `-sin^-1[sin(- pi/6)]`

= `-(- pi/6)`

= `pi/(6)`.

shaalaa.com
Fundamental Theorem of Integral Calculus
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Exercise 4.2 [पृष्ठ १७१]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 4 Definite Integration
Exercise 4.2 | Q 1.12 | पृष्ठ १७१

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