Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{Let I }=\int_0^1 x\log\left( 1 + 2x \right)dx\]
Applying integration by parts, we have
\[I = \log\left( 1 + 2x \right)\frac{x^2}{2}_0^1 - \int_0^1 \left( \frac{2}{1 + 2x} \right) \times \frac{x^2}{2}dx\]
\[ = \frac{1}{2}\left( \log3 - 0 \right) - \int_0^1 \frac{x^2}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{4 x^2 - 1 + 1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \frac{\left( 2x + 1 \right)\left( 2x - 1 \right)}{1 + 2x}dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]
\[ = \frac{1}{2}\log3 - \frac{1}{4} \int_0^1 \left( 2x - 1 \right)dx - \frac{1}{4} \int_0^1 \frac{1}{1 + 2x}dx\]
\[= \left.\frac{1}{2}\log3 - \frac{1}{4} \times \frac{\left( 2x - 1 \right)^2}{2 \times 2}\right|_0^1 - \left.\frac{1}{4} \times \frac{\log\left( 1 + 2x \right)}{2}\right|_0^1 \]
\[ = \frac{1}{2}\log3 - \frac{1}{16}\left( 1 - 1 \right) - \frac{1}{8}\left( \log3 - \log1 \right)\]
\[ = \frac{1}{2}\log3 - 0 - \frac{1}{8}\log3 ....................\left( \log1 = 0 \right)\]
\[ = \frac{3}{8}\log3\]
APPEARS IN
संबंधित प्रश्न
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
