हिंदी

1 ∫ 0 Log ( 1 X − 1 ) D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx\]

 

योग
Advertisements

उत्तर

\[Let\ I = \int_0^1 \log\left( \frac{1}{x} - 1 \right) d x ...............(1)\]
\[ = \int_0^1 \log\left( \frac{1}{1 - x} - 1 \right) d x ...............\left[\text{Using }\int_0^a f(x) dx = \int_0^a f(a - x) dx \right]\]
\[ I = \int_0^1 \log\left( \frac{x}{1 - x} \right) dx ...............(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^1 \log\left( \frac{1 - x}{x} \right) + \log\left( \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log\left( \frac{1 - x}{x} \times \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log1 dx \]
\[ = 0\]
\[Hence\ I = 0\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.5 | Q 34 | पृष्ठ ९५

संबंधित प्रश्न

\[\int\limits_0^1 \frac{x}{x + 1} dx\]

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

\[\int\limits_0^{\pi/2} \cos^2 x\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

\[\int\limits_1^e \frac{e^x}{x} \left( 1 + x \log x \right) dx\]

\[\int\limits_1^e \frac{\log x}{x} dx\]

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]

\[\int\limits_0^1 \tan^{- 1} x\ dx\]

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_0^2 \left( x^2 + x \right) dx\]

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^2} dx\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


\[\int\limits_1^e \log x\ dx =\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

\[\int\limits_0^4 x\sqrt{4 - x} dx\]


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{2\pi} \cos^7 x dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


\[\int\limits_2^3 e^{- x} dx\]


\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×