Advertisements
Advertisements
प्रश्न
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
योग
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
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/4} \sec x dx\]
\[\int\limits_0^{\pi/2} \cos^2 x\ dx\]
\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]
\[\int\limits_0^{\pi/2} \cos x\ dx\]
\[\int\limits_2^3 \frac{1}{x}dx\]
Evaluate :
\[\int\limits_2^3 3^x dx .\]
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^4 x dx\]
