Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_1^2 \left( \frac{x - 1}{x^2} \right) e^x\ d\ x . Then, \]
\[I = \int_1^2 \left( \frac{e^x}{x} - \frac{e^x}{x^2} \right) dx\]
\[ \Rightarrow I = \int_1^2 \frac{e^x}{x} dx - \int_1^2 \frac{e^x}{x^2} dx\]
\[\text{Integrating first term by parts}\]
\[I = \left\{ \left[ \frac{e^x}{x} \right]_1^2 - \int_1^2 \frac{- 1}{x^2} e^x dx \right\} - \int_1^2 \frac{e^x}{x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^x}{x} \right]_1^2 + \int_1^2 \frac{e^x}{x^2} dx - \int_1^2 \frac{e^x}{x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^x}{x} \right]_1^2 \]
\[ \Rightarrow I = \frac{e^2}{2} - e\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Solve each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
