Advertisements
Advertisements
Question
Evaluate the following:
`Γ (9/2)`
Sum
Advertisements
Solution
`Γ (9/2) = (9 /2 - 1) Γ(9/2 - 1)`
= `7/2 Γ 7/2`
= `7/2 5/2 Γ 5/2`
= `7/2 5/2 3/2 Γ 3/2`
= `7/2 5/2 3/2 1/2 Γ (1/2)`
= `(105sqrt(pi))/16`
shaalaa.com
Is there an error in this question or solution?
Chapter 2: Integral Calculus – 1 - Exercise 2.10 [Page 51]
APPEARS IN
RELATED QUESTIONS
\[\int_0^1 x\log\left( 1 + 2x \right)dx\]
\[\int\limits_0^2 x\sqrt{x + 2}\ dx\]
\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]
\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]
\[\int\limits_a^b \cos\ x\ dx\]
\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
