Calculus an integral part of life

The functional series is uniformly convergent. So we can apply integration first and then do the summation. Then it becomes the infinite summation of k/k! . Which is nothing but e .

We know that $\displaystyle \sum_{i=0}^\infty$ $\frac{x^k}{k!}$ = $e^x$

Now, Multiplying by x on both sides and Differentiating twice with respect to x, we get

$\displaystyle \sum_{i=0}^\infty$ $\frac{x^{k-1}}{k!}$ k(k+1)= $e^x$ $(x+2)$

Multiplying by x again gives us the required sum $f_k(x)$

$f_k(x)$ = $e^x$ $x(x+2)$

$\int\limits_0^1$ $f_k(x)$ $dx$ = $\int\limits_0^1$ $e^x$ $x(x+2)$ $dx$ = $e$