Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_1^2 \log\ x\ d\ x\ . Then, \]
\[I = \int_1^2 1 \log x\ d\ x\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 \frac{1}{x} x\ d\ x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 d x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \left[ x \right]_1^2 \]
\[ \Rightarrow I = 2 \log 2 - 2 + 1\]
\[ \Rightarrow I = 2 \log 2 - 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
Evaluate each of the following integral:
Evaluate each of the following integral:
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Choose the correct alternative:
Γ(1) is
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
