Advertisements
Advertisements
प्रश्न
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Advertisements
उत्तर
Let f(x = `log (2 - x)/(2 + x))`
f(– x) = `log ((2 - (- x))/(2 + (– x)))`
= `log ((2 + x)/(2 - x))`
= `log ((2 - x)/(2 + x))^-1`
= `- log ((2 - x)/(2 + x))`
⇒ (fx) = – f(x)
∴ f(x) is an odd function
∴ `int_(-1)^1 log ((2 - x)/(2 + x)) "d"x` = 0
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
