find :
\frac{29\int_{0}^{1}{\left(1-x^4 \right)^{7}}\mathrm{dx}}{4\int_{0}^{1}{\left(1-x^4 \right)^{6}}\mathrm{dx}}
1Nishant sir ... can't it be done in some other way like:
(1-x4)7 can be broken down to (1-x4)6(1-x4)
The expr. boils down to 29/4 * [ 1 - ∫x4(1-x4)6dx∫(1-x4)6dx ]
then (1-x4)6 can be binomially expanded in both numerator and denominator and integrated [ higher terms neglected. ]
The solution might come in a better form...
1@arka wat bhaiya was telling was a hilarious stuff
and its pretty big if u go for binomial expansion
u may use reduction
1By the way ... Sonne can u please tell me of any easier way with lesser steps?????
1i didnt get wat u mean by neglecting higher powers here ? can u explain it again
http://latex.codecogs.com/gif.latex?\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=x.(1-x^4)^7|_0^1+\int_{0}^{1}28x^4(1-x^4)^6\mathrm{dx}\\%20\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=\int_{0}^{1}28(1-x^4-1)(1-x^4)^6)\mathrm{dx}\\%20\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=\int_{0}^{1}28\left%20((1-x^4)^6)-(1-x^4)^7%20\right%20)\mathrm{dx}\\%20\frac{29\int_{0}^{1}(1-x^4)^7\mathrm{dx}}{4\int_{0}^{1}(1-x^4)^6\mathrm{dx}}=7
1Hey ... what's the use of pinked post here???????
Nobody has given the answer yet!!!!!
1Let I = 0 ∫ 1 ( 1 - x 4 ) 7 dx
J = 0 ∫ 1 ( 1 - x 4 ) 6 dx
Let Us Part - Integrate " I " .
I = [ ( x ) ( 1 - x 4 ) 7 ] 0 1 - 0 ∫ 1 ( 1 - x 4 ) 6 { - 7 . 4 x 3 } ( x ) dx
= 0 - 0 ∫ 1 ( 1 - x 4 ) 6 ( 28 ) ( 1 - x 4 - 1 ) dx
= - ( 28 ) 0 ∫ 1 ( 1 - x 4 ) 7 dx + ( 28 ) 0 ∫ 1 ( 1 - x 4 ) 6 dx
= - 28 I + 28 J
So , I = - 28 I + 28 J .
Or , J = 2928 I
Given Integral K = 294 IJ = 284 = 7
P . S --- Calculation Mistake Corrected .
1@ricky u made a calculation mistake in
calculating the derivative of
(1-x4)7
1I = \frac{29\int_{0}^{1}( 1- x^4)^7dx}{4\int_{0}^{1}(1-x^4)^6 dx}
\frac{1}{I} = \frac{4\int_{0}^{1}(1-x^4)^6 dx}{29\int_{0}^{1}( 1- x^4)^7dx}
\frac{29}{4I} = \frac{\int_{0}^{1}(1-x^4)^6 dx}{\int_{0}^{1}( 1- x^4)^7dx}
\int_{0}^{1}( 1- x^4)^7 . 1 dx \rightarrow x(1-x^4)^7 |_{0}^{1} - \int_{0}^{1}7 ( 1- x^4)^6( -4x^3)(x)
0 +28 \int_{0}^{1} ( 1- x^4)^6(x^4)
28 \int_{0}^{1} ( 1- x^4)^6(x^4)
\frac{29. 28}{4I } = \frac{\int_{0}^{1}(1-x^4)^6}{\int_{0}^{1}(1-x^4)^6)(x^4+1-1)}
\frac{29. 28}{4I } = \frac{\int_{0}^{1}(1-x^4)^6}{\int_{0}^{1}(1-x^4)^6 -\int_{0}^{1}(1-x^4)^7}
\frac{29. 28}{4I } = \frac{1}{\frac{\int_{0}^{1}(1-x^4)^6 -\int_{0}^{1}(1-x^4)^7}{\int_{0}^{1}(1-x^4)^6}}
\frac{29. 28}{4I } = \frac{1}{1 -\frac{\int_{0}^{1}(1-x^4)^7}{\int_{0}^{1}(1-x^4)^6}}
\frac{29. 28}{4I } = \frac{1}{1 -\frac{4I}{29}}
\frac{29. 28}{4I } = \frac{29}{29 -{4I}}
\frac{ 7}{I } = \frac{1}{29 -{4I}}
(29)(7)- 28 I = I
\not{29}.7 = \not{29}I
I = \boxed 7
62Good work both u.. sonne and soumi [1]
Sonne's hidden post ;)
\int_{0}^{1}\left ( 1-x^4 \right )^7\mathrm{dx}=x.(1-x^4)^7|_0^1+\int_{0}^{1}28x^4(1-x^4)^6\mathrm{dx}\\ \int_{0}^{1}\left ( 1-x^4 \right )^7\mathrm{dx}=\int_{0}^{1}28(1-x^4-1)(1-x^4)^6)\mathrm{dx}\\ \int_{0}^{1}\left ( 1-x^4 \right )^7\mathrm{dx}=\int_{0}^{1}28\left ((1-x^4)^6)-(1-x^4)^7 \right )\mathrm{dx}\\ \frac{29\int_{0}^{1}(1-x^4)^7\mathrm{dx}}{4\int_{0}^{1}(1-x^4)^6\mathrm{dx}}=7
