Let f:(0, infty) rightarrow mathbb{R} and F(x | vedclass

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(C) Given $F(x) = \int_0^x t f(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x f(x)$.Given $F(x^2) = x^4 + x^5$. Let $u = x^2$,then $F(u) =

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$219$

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(C) Given $F(x) = \int_0^x t f(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x f(x)$.Given $F(x^2) = x^4 + x^5$. Let $u = x^2$,then $F(u) = u^2 + u^{5/2}$.Differentiating with respect to $u$,we get $F'(u) = 2u + \frac{5}{2} u^{3/2}$.Since $F'(u) = u f(u)$,we have $u f(u) = 2u + \frac{5}{2} u^{3/2}$.Dividing by $u$,we get $f(u) = 2 + \frac{5}{2} u^{1/2}$.We need to find $\sum_{r=1}^{12} f(r^2)$. Substituting $u = r^2$,we get $f(r^2) = 2 + \frac{5}{2} (r^2)^{1/2} = 2 + \frac{5}{2} r$.Thus,$\sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} (2 + \frac{5}{2} r) = \sum_{r=1}^{12} 2 + \frac{5}{2} \sum_{r=1}^{12} r$.$= 2(12) + \frac{5}{2} \left( \frac{12 	imes 13}{2} ight) = 24 + \frac{5}{2} (78) = 24 + 5(39) = 24 + 195 = 219$.

Let $f:(0, \infty) \rightarrow \mathbb{R}$ and $F(x)=\int_0^x t f(t) d t$. If $F(x^2)=x^4+x^5$,then $\sum_{r=1}^{12} f(r^2)$ is equal to :

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