Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 45334 by Meritguide1234 last updated on 12/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
	$using gamma beta function ∫_0 ^1 t^(m−1) (1−t)^(n−1) dt=β(m,n)=((⌈(m)⌈(n))/(⌈(m+n))) ⌈(n+1)=n! I_1 =∫_0 ^1 x^(1004) (1−x)^(1004) dx=((⌈(1005)⌈(1005))/(⌈(1005+1005)))←I_1 here m−1=1004 m=1005 n−1=1004 n=1005 I_2 = ∫_0 ^1 x^(1004) (1−x^(2010) )^(1004) dx y=x^(2010) dy=2010 x^(2009) dx x=y^(1/(2010)) dx=(dy/(2010))×(1/((y)^((2009)/(2010)) )) ∫_0 ^1 (y)^((1004)/(2010)) (1−y)^(1004) ×(dy/(2010))×(1/y^((2009)/(2010)) ) =(1/(2010))∫_0 ^1 (y)^((1004−2009)/(2010)) (1−y)^(1004) dy =(1/(2010))∫_0 ^1 y^((−1)/2) (1−y)^(1004) dy =(1/(2010))∫_0 ^1 y^((1/2)−1) (1−y)^(1005−1) dy =(1/(2010))×((⌈((1/2))×⌈(1005))/(⌈(1005+(1/2))))←I_2 ans is (2^(2010) /(1005))×(I_1 /I_2 ) =(2^(2010) /(1005))×((⌈(1005)×⌈(1005))/(⌈(2010)))×2010×((⌈(1005+(1/2)))/(⌈((1/2))×⌈(1005))) =(2^(2010) /(1005))×2010×{⌈(1005)×⌈(1005+(1/2))×2^(2×1005−1) }×(1/2^(2×1005−1) )×(1/(⌈(2010)))×(1/(⌈((1/2)))) =2^(2011) ×{(√π) ×⌈(2010)}×(1/2^(2009) )×(1/((√π) ×⌈(2010))) =2^(2011−2009) =2^2 =4 so ans is 4 formula 2^(2x−1) ⌈(x)⌈(x+(1/2))=(√π) ×⌈(2x) pls check...$
	$u s i n g g a m m a b e t a f u n c t i o n$ $\int_{0}^{1} t^{m - 1} {(1 - t)}^{n - 1} d t = β (m, n) = \frac{⌈ (m) ⌈ (n)}{⌈ (m + n)}$ $⌈ (n + 1) = n!$ $I_{1} = \int_{0}^{1} x^{1004} {(1 - x)}^{1004} d x = \frac{⌈ (1005) ⌈ (1005)}{⌈ (1005 + 1005)} \leftarrow I_{1}$ $h e r e m - 1 = 1004 m = 1005$ $n - 1 = 1004 n = 1005$ $I_{2} =$ $\int_{0}^{1} x^{1004} {(1 - x^{2010})}^{1004} d x$ $y = x^{2010} d y = 2010 x^{2009} d x$ $x = y^{\frac{1}{2010}}$ $d x = \frac{d y}{2010} \times \frac{1}{{(y)}^{\frac{2009}{2010}}}$ $\int_{0}^{1} {(y)}^{\frac{1004}{2010}} {(1 - y)}^{1004} \times \frac{d y}{2010} \times \frac{1}{y^{\frac{2009}{2010}}}$ $= \frac{1}{2010} \int_{0}^{1} {(y)}^{\frac{1004 - 2009}{2010}} {(1 - y)}^{1004} d y$ $= \frac{1}{2010} \int_{0}^{1} y^{\frac{- 1}{2}} {(1 - y)}^{1004} d y$ $= \frac{1}{2010} \int_{0}^{1} y^{\frac{1}{2} - 1} {(1 - y)}^{1005 - 1} d y$ $= \frac{1}{2010} \times \frac{⌈ (\frac{1}{2}) \times ⌈ (1005)}{⌈ (1005 + \frac{1}{2})} \leftarrow I_{2}$ $a n s i s \frac{2^{2010}}{1005} \times \frac{I_{1}}{I_{2}}$ $= \frac{2^{2010}}{1005} \times \frac{⌈ (1005) \times ⌈ (1005)}{⌈ (2010)} \times 2010 \times \frac{⌈ (1005 + \frac{1}{2})}{⌈ (\frac{1}{2}) \times ⌈ (1005)}$ $= \frac{2^{2010}}{1005} \times 2010 \times {⌈ (1005) \times ⌈ (1005 + \frac{1}{2}) \times 2^{2 \times 1005 - 1}} \times \frac{1}{2^{2 \times 1005 - 1}} \times \frac{1}{⌈ (2010)} \times \frac{1}{⌈ (\frac{1}{2})}$ $= 2^{2011} \times {\sqrt{π} \times ⌈ (2010)} \times \frac{1}{2^{2009}} \times \frac{1}{\sqrt{π} \times ⌈ (2010)}$ $= 2^{2011 - 2009} = 2^{2} = 4 s o a n s i s 4$ $f o r m u l a$ $2^{2 x - 1} ⌈ (x) ⌈ (x + \frac{1}{2}) = \sqrt{π} \times ⌈ (2 x)$ $p l s c h e c k . . .$
	Commented by Meritguide1234 last updated on 12/Oct/18

	$y e s g o o d s o l u t i o n$
	Commented by Meritguide1234 last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$y o u r p o s t a r e r e a l l y p r s i s e w o r t h y a n d t o u g h . . .$
	Commented by Meritguide1234 last updated on 12/Oct/18

	$m a t h i s n o t t o u g h . . . j u s t e a s y t o u n d e r s t a n d a b l e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum