Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
योग
Advertisements
उत्तर
`int_0^oo "e"^(- x/2) x^5 "d"x = (5!)/(1/2)^(5+ 1)`
= `(5!)/(1/2)^6`
= (26)5!
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]
\[\int\limits_0^{\pi/2} \cos^3 x\ dx\]
\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]
\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]
Evaluate the following integral:
\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]
\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
