Let f:(0, ∞) → R and F(x) = ∫ tf(t)dt; t ∈ [0, x]. F(x^2) = {x}^{4}+{x}^{5},

Question

Let \(f:(0, \infty) \rightarrow R\) and \(F(x)=\int\limits_{0}^{x} t f(t) d t\). If \(F\left(x^{2}\right)={x}^{4}+{x}^{5}\), then \(\sum\limits_{{r}=1}^{12} {f}\left({r}^{2}\right) \) is equal to _____.

Answer 1

Correct answer: 219

\( F(x)=\int\limits_{0}^{x} t \cdot f(t) d t\)

\( F^{1}(x)=x f(x)\)

\(\text {Given } \mathrm{F}\left(\mathrm{x}^{2}\right)=\mathrm{x}^{4}+\mathrm{x}^{5}, \quad \text { let } \mathrm{x}^{2}=\mathrm{t}\)

\(\mathrm{F}(\mathrm{t})=\mathrm{t}^{2}+\mathrm{t}^{5 / 2}\)

\(F^{\prime}(t)=2 t+5 / 2 t^{3 / 2}\)

\( t \cdot f(t)=2 t+5 / 2 t^{3 / 2}\)

\(f(t)=2+5 / 2 r^{1 / 2}\)

\( \sum\limits_{r=1}^{12} f\left(r^{2}\right)=\sum\limits_{r=1}^{12} 2+\frac{5}{2} r \)

\(=24+5 / 2\left[\frac{12(13)}{2}\right]\)

\(=219\)