Advertisements
Advertisements
प्रश्न
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
विकल्प
- \[\frac{a + b}{2} \int\limits_a^b f\left( b - x \right) dx\]
- \[\frac{a + b}{2} \int\limits_a^b f\left( b + x \right) dx\]
- \[\frac{b - a}{2} \int\limits_a^b f\left( x \right) dx\]
- \[\frac{b + a}{2} \int\limits_a^b f\left( x \right) dx\]
Advertisements
उत्तर
\[Let\, I = \int_a^b x f\left( x \right) d x .............(1)\]
\[ = \int_a^b \left( a + b - x \right) f\left( a + b - x \right) d x\]
\[ = \int_a^b \left( a + b - x \right) f\left( x \right) dx ...............(2)\]
\[ \text{Adding (1) and (2)}\]
\[2I = \int_a^b \left( x + a + b - x \right) f\left( x \right) d x\]
\[ = \left( a + b \right) \int_a^b f\left( x \right) d x \]
\[Hence\ I = \frac{a + b}{2} \int_a^b f\left( x \right) d x\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
