मराठी

The Derivative of F ( X ) = X 3 ∫ X 2 1 Log E T D T , ( X > 0 ) , Is1 3 Ln X,1 3 Ln X − 1 2 Ln X,,(Ln X)−1 X (X − 1),3 X 2 Ln X

Advertisements
Advertisements

प्रश्न

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

 

पर्याय

  • \[\frac{1}{3 \ln x}\]
  • \[\frac{1}{3 \ln x} - \frac{1}{2 \ln x}\]
  • (ln x)−1 x (x − 1)

  • \[\frac{3 x^2}{\ln x}\]
MCQ
Advertisements

उत्तर

(ln x)−1 x (x − 1)

Using Newton Leibnitz formula

\[f' (x) = \frac{1}{\log_e x^3}(3 x^2 ) - \frac{1}{\log_e x^2}(2x) \]

\[= \frac{3 x^2}{3\ln x}- \frac{2x}{2\ln x} \]

\[= \frac{x^2}{\ln x} - \frac{x}{\ln x} \]

\[= \frac{1}{\ln x}x(x - 1) \]

\[= {(\ln x)}^{- 1} x(x - 1)\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ ११९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 26 | पृष्ठ ११९

संबंधित प्रश्‍न

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

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

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

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

\[\int\limits_0^2 x\sqrt{2 - x} dx\]

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

\[\int\limits_0^4 \left( x + e^{2x} \right) dx\]

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

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

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

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

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


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

\[\int\limits_0^{\pi/2} x \sin x\ dx\]  is equal to

Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


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


\[\int\limits_{- 1}^1 e^{2x} dx\]


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×