Integral `int_a^b f(x)dx` can be represented as a limit of a sum ofinfinite series `int_a^b f(x) dx=lim_(n-oo) sum_(r=na+C)^(nb+-c) 1/n f(r/n)` where, `na + c leq nb+c,r,n in N,c inR` and any limit of sumof series of same form can be changed to definite integral by replacing

Integral int_(a)^(b)f(x)dx can be represented as a limit of a sum ofinfinite series int_(a)^(b)f(x)dx=lim_(n-oo)sum_(r=na+C)^(nb+-c)(1)/(n)f((r)/(n)) where na+c<=nb+c,r,n in N,c in R and any limit of sumof series of same form can be changed to definite integral by replacing

Definite integration as the limit of a sum : lim_(ntooo)(1)/(n)sum_(r=n+1)^(2n)log(1+(r)/(n))=.............

If sum_(r=1)^(n) r(r+1) = ((n+a)(n+b)(n+c))/(3) , where a gt b gt c , then

The sum of the series sum_(r=0) ^(n) ""^(2n)C_(r), is

The sum of the series sum_(r=0) ^(n) ""^(2n)C_(r), is

If I_n = int sin^n x dx then nI_n - (n - 1)I_(n-2) = f(x)+c where f(x) =

If f(a+x)=f(x), then prove that int_(a)^(na)f(x)dx=(n-1)int_(0)^(a)f(x)dx where a>0 and n in N.

Suppose in the definite integral int_a^b f(x) dx the upper limit b->oo, then to obtain the value of int_a^bf(x) dx , we may say that int_a^bf(x)dx=lim_(k->oo)int_a^k dx, where k > a. if f(x)->oo as x ->a or x->b, then the value of definite integral int_a^bf(x)dx is lim_(h->0) int_(a+h)^b f(x) dx. If this limit the value of the limit is defined as the value of integral. This should be noted that f (x)should not have any other discontinuity in [a, b] otherwise this will lead to errorous solution.

If f(n)=sum_(s=1)^(n)sum_(r=s)^(n)C_(r)^(r)C_(s), then f(3)=

Definite integration as the limit of a sum : lim_(ntooo)sum_(r=1)^(n)(1)/(n)e^(r/(n))=.............