मराठी

Evaluate Each of the Following Integral: ∫ E 2 E 1 X Log X D X - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]
बेरीज
Advertisements

उत्तर

\[\int_e^{e^2} \frac{1}{x\log x}dx\]
\[ = \int_e^{e^2} \frac{\frac{1}{x}}{\log x}dx\]
\[ = \left.\log\left( \log x \right)\right|_e^{e^2} ...............\left[ \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log f\left( x \right) + C \right]\]
\[ = \log\left( \log e^2 \right) - \log\left( \log e \right)\]
\[ = \log\left( 2\log e \right) - \log\left( \log e \right) \]
\[ = \log2 - \log1 ................\left( \log e  = 1 \right)\]
\[ = \log2 - 0\]
\[ = \log2\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Very Short Answers | Q 28 | पृष्ठ ११५

संबंधित प्रश्‍न

\[\int\limits_1^2 \log\ x\ dx\]

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

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

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

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

\[\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx\]

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

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

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]

 


\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

\[\int\limits_0^\pi x \log \sin x\ dx\]

If f(2a − x) = −f(x), prove that

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

\[\int\limits_0^2 \left( x + 3 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

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

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

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


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


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


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


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following:

`Γ (9/2)`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×