Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{1 - x}{1 + x} d\ x\ . Then, \]
\[I = \int_0^1 \left( \frac{1}{1 + x} - \frac{1 + x - 1}{1 + x} \right) d x\]
\[I = \int_0^1 \left( \frac{1}{1 + x} - 1 + \frac{x}{1 + x} \right) d x\]
\[ \Rightarrow I = \left[ \log \left( 1 + x \right) - x + \log \left( 1 + x \right) \right]_0^1 \]
\[ \Rightarrow I = \left( \log 2 - 1 + \log 2 \right) - \left( \log 1 - 0 + \log 1 \right)\]
\[ = 2 \log 2 - 1\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
