1(a-x)(b-x)(c-x)......(x-x)....(z-x)=0
so answer is K
Solve :- \int \left(a-x \right)\left(b-x \right)\left(c-x \right)...............\left(z-x \right)dx
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1(a-x)(b-x)....(z-x) = (x-a)(x-b)...(x-z)
= x26-(a+b+..+z)x25+...+(abc..z)
So, ∫(a-x)(b-x)....(z-x)dx= ∫x26-(a+b+..+z)x25+...+(abc..z)dx
=x27/27 -(a+b+..+z)x26/26+...+(ab..z)x +C
1i had already askd this integration...................
and ans is a constant
1No the answer is not K you boys are wrong wait I will post the comment lets see whether anybody else can solve it or not
1Don't be a fool u have taken the constant as "k" and i have taken the constant as "c" Ha Ha Ha Ha HA Hi hi hi ho
9itll be better to post it in humor/fun
so that metal wudnt have wasted his genious in doing it correctly :P
1Boys, actually it can be done in a more simple way as Akshay has tried to describe
(a-x)(b-x).............(x-x)(y-x)(z-x)=0
Hence,∫(a-x)(b-x).............(z-x)dx
=0+c
=c
Now somebody may take "k" in place of "c"that doesn't matter but I appreciate
METAL