Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_1^e \frac{e^x}{x}\left( 1 + x \log x \right)\ d\ x\ . Then, \]
\[I = \int_1^e \left( \frac{e^x}{x} + e^x \log x \right) dx\]
\[ \Rightarrow I = \int_1^e \frac{e^x}{x} dx + \int_1^e e^x \log x\ d\ x\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left[ \log x e^x \right]_1^e - \int_1^e e^x \log x d x + \int_1^e e^x \log\ x\ d\ x\]
\[ \Rightarrow I = \left( \log e \right) e^e - 0\]
\[ \Rightarrow I = e^e\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Prove that:
Evaluate each of the following integral:
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
