Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{x}{x + 1} d x . Then, \]
\[I = \int_0^1 1 - \frac{1}{x + 1} d x\]
\[ \Rightarrow I = \left[ x - \log \left( x + 1 \right) \right]_0^1 \]
\[ \Rightarrow I = 1 - \log 2 - (0 - \log 1)\]
\[ \Rightarrow I = \log e - \log 2\]
\[ \Rightarrow I = \log \frac{e}{2}\]
APPEARS IN
संबंधित प्रश्न
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
