Advertisements
Advertisements
प्रश्न
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
योग
Advertisements
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]
\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]
\[\int\limits_0^\infty \frac{\log x}{1 + x^2} dx\]
\[\int\limits_1^3 \left( 3x - 2 \right) dx\]
\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]
Solve each of the following integral:
\[\int_2^4 \frac{x}{x^2 + 1}dx\]
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
