Question

Find i1/i2

Answer 1

\[\LARGE I_1=29\int\limits_{0}^{1} (1-x^4)^7dx\]
\[\LARGE I_2=4\int\limits_{0}^{1} (1-x^4)^6dx\]

Answer 2

@zepdrix @ganeshie8 @experimentX @amistre64

Answer 3

Attempt: Ill try to write i1 in terms of i2,anyone can post if they got better methods :)
\[Applying~~integration~~by~~parts\]
\[\large I_1=x(1-x^4)^7]_{0}^{1}+28\int\limits\limits\limits_{0}^{1}((1-x)^4.x^4)dx\]

Answer 4

\[\large I_1=0+28\int\limits\limits\limits\limits_{0}^{1}((1-x)^4.x^4)dx\]

Answer 5

do you know a binomial expansion.

Answer 6

Correction
\[\large I_1=0+28\int\limits\limits\limits\limits\limits_{0}^{1}((1-x^4)^6.x^4)dx\]

Answer 7

and yes

Answer 8

use binomial theorem to expand it and integrate term by term.

Answer 9

....complex

Answer 10

no complex ... this is simple.
\[ 29\int\limits_{0}^{1} (1-x^4)^7dx = 29 \int_0^1 \sum_{k=0}^7 (-1)^k \binom{7}{k}x^{4k}dx = \sum_{k=0}^7 (-1)^k \binom{7}{k} \frac{1}{4k + 1}\]

Answer 11

add *29 to that last one ... and solve the other same way ... then just put up in your calculator and add up to find the sum.

Answer 12

internet is pretty slow here ... plug this thing into wolf and see if they are equal or not
Integrate[(1 - x^4)^7, {x, 0, 1}]
Sum[(-1)^k Binomial[7, k]/(4 k + 1), {k, 0, 7}]