Advertisements
Advertisements
प्रश्न
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
बेरीज
उत्तर
`int_0^oo x^2 "e"^(-2x) "d"x = int_0^oo x^"n""e"^(-"a"x) "d"x`
= `("n"!)/("a"^("n" + 1))`
Where n = 2
a = 2
So `int_0^oo "f"(x) "d"x = (2!)/2^3`
= `2/8`
= `1/4`
shaalaa.com
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]
\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]
\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]
\[\int_0^1 | x\sin \pi x | dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]
\[\int\limits_1^2 \log_e \left[ x \right] dx .\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals
\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\]
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
Choose the correct alternative:
Γ(n) is
