Evaluate `int_(0)^(3)(x^2-6x+5)dx`

Answer

B

Explanation

a. We know that
`.^(m)C_(r)+.^(m)C_(r-1)xx.^(n)C_(1)+.^(m)C_(r-2)xx.^(n)C_(2)+"….."+.^(n)C_(r)= .^(m+n)C_(r)`
So, if `m, n gt r` and `m + n lt r`, then
`.^(m)C_(r)+.^(m)C_(r-1)xx.^(n)C_(1)+.^(m)C_(r-2)xx.^(n)C_(2)+"….."+.^(n)C_(r)=0`
b. if `m, n gt r` then
`.^(m)C_(r)+.^(m)C_(r-1)xx.^(n)C_(1)+.^(m)C_(r-2)xx.^(n)C_(2)+"....."+.^(n)C_(r)=.^(m+n)C_(r)`
c. If `m, n lt r lt m + n` then
`.^(m)C_(m) x .^(n)C_(r-m) + .^(m)C_(m-1)xx.^(n)C_(r-m+1)`
`+ .^(m)C_(m-2)xx.^(n)C_(r-m+2)+"...."+.^(m)C_(r-n)xx.^(n)C_(n)=.^(m+n)C_(r)`
d. if `m lt r lt n`
`.^(m)C_(m) xx .^(n)C_(r-m)+.^(m)C_(m-1)xx.^(n)C_(r-m+1) +.^(m)C_(m-2)xx.^(n)C_(r-m+2)+"....."+.^(n)C_(r)=.^(m+n)C_(r)`.