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Evaluate the following : ∫0a1a2+ax-x2⋅dx - Mathematics and Statistics

Sum

Evaluate the following : `int_0^a 1/(a^2 + ax - x^2)*dx`

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Solution

Let I = `int_0^a 1/(a^2 + ax - x^2)*dx`

a2 + ax – x2 = `a^2 - (x^2 - ax + a^2/4) + a^2/(4)`

= `(5a^2)/(4) - (x - a/2)^2`

= `(sqrt(5a)/2)^2 - (x - a/2)^2`

∴ I = `int_0^a dx/(((sqrt(5)a)/2)^2 - (x - a/2)^2)`

= `(1)/((2 xx sqrt(5)a)/2)*[log|((sqrt(5)a)/2 + x - a/2)/((sqrt(5)a)/(2) - x + a/2)|]_0^a`

= `(1)/(sqrt(5)a)[log|((sqrt(5)a)/2 + a - a/2)/((sqrt(5)a)/(2) - a + a/2)| - log |((sqrt(5)a)/2 - a/2)/((sqrt(5)a)/(2) + a/2)|]`

= `(1)/(sqrt(5)a)[log |(sqrt(5)/2 + 1/2)/(sqrt(5)/2 - 1/2)| - log |(sqrt(5)/2 - 1/2)/(sqrt(5)/2 + 1/2)|]`

= `(1)/(sqrt(5)a)[log|((sqrt(5) + 1)/(sqrt(5) - 1))|- log|((sqrt(5) - 1)/(sqrt(5) + 1))|]`

= `(1)/(sqrt(5)a) log|(sqrt(5) + 1)/(sqrt(5) - 1) xx (sqrt(5) + 1)/(sqrt(5) - 1)|`

= `(1)/(sqrt(5)a) log [((sqrt(5) + 1)/(sqrt(5) - 1))^2]`

= `(1)/(sqrt(5)a) log |(5 + 1 + 2sqrt(5))/(5 + 1  - 2sqrt(5))|`

= `(1)/(sqrt(5)a) log  (6 + 2sqrt(5))/(6 - 2sqrt(5))`

= `(1)/(sqrt(5)a) log|(6 + 2sqrt(5))/(6 - 2sqrt(5)) xx (6 + 2sqrt(5))/(6 + 2sqrt(5))|`

= `(1)/(sqrt(5)a) log|(36 + 20 + 24sqrt(5))/(36 - 20)|`

= `(1)/(sqrt(5)a) log |(56 + 24sqrt(5))/(16)|`

= `(1)/(sqrt(5)a) log|(7 + 3sqrt(5))/(2)|`.

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 3.03 | Page 176
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