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
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Write the coefficient a, b, c of which the value of the integral
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Choose the correct alternative:
Γ(1) is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
