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Sum

Solve the following : `int_0^1 (1)/(2x - 3)*dx`

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#### Solution

Let I = `int_0^1 (1)/(2x - 3)*dx`

Put 2x – 3 =t

∴ 2·dx = dt

∴ dx = `"dt"/(2)`

When x = 0t = 2(0) – 3 = – 3

When x = 1, t = 2(1) – 3 = – 1

∴ I = `int_(-3)^(-1) (1)/"t"*"dt"/(2)`

= `(1)/(2) int_(-3)^(-1) "dt"/"t"`

= `(1)/(2)[log |"t"|]_(-3)^(-1)`

= `(1)/(2)[log|-1| - log|-3|]`

= `(1)/(2)(log 1 - log 3)`

= `(1)/(2)(0 - log 3)`

∴ I = `-(1)/(2) log 3`.

Concept: Fundamental Theorem of Integral Calculus

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