Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2 x^2} \right) d\ x\ . Then, \]
\[I = \int_1^2 e^{2x} \frac{1}{x} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{e^{2x}}{2x} \right]_1^2 - \int_1^2 - e^{2x} \frac{1}{2 x^2} \right\} - \int_1^2 e^{2x} \frac{1}{2 x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^{2x}}{2x} \right]_1^2 \]
\[ \Rightarrow I = \frac{e^4}{4} - \frac{e^2}{2}\]
\[ \Rightarrow I = \frac{e^4 - 2 e^2}{4}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
