Let f:(0,∞)→R and letF(x)=∫0x f(t)dtIf Fx2=x2(1+x) then f (4) equals

F(x)=∫0x f(t)dt⇒F′(x)=f(x)∴f(4)=F′(2) Also, 2xF′x2=2x(1+x)+x2⇒F′x2=1+3x/2∴f(4)=F′22=1+3=4

Let f:(0,∞)→R and letF(x)=∫0x f(t)dtIf Fx2=x2(1+x) then f (4) equals

F(x)=∫0x f(t)dt⇒F′(x)=f(x)∴f(4)=F′(2) Also, 2xF′x2=2x(1+x)+x2⇒F′x2=1+3x/2∴f(4)=F′22=1+3=4

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Let $f : (0, \infty) \to R$ and let
$F (x) = \int_{0}^{x} f (t) d t$
If $F (x^{2}) = x^{2} (1 + x)$ then $f (4)$ equals

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$\frac{5}{4}$

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Detailed Solution

$\begin{aligned} F (x) = \int_{0}^{x} f (t) d t \\ \Rightarrow F^{'} (x) = f (x) \\ ∴ f (4) = F^{'} (2) \end{aligned}$

$\begin{aligned} Also, 2 x F^{'} (x^{2}) = 2 x (1 + x) + x^{2} \\ \Rightarrow F^{'} (x^{2}) = 1 + 3 x / 2 \end{aligned}$

$∴ f (4) = F^{'} (2^{2}) = 1 + 3 = 4$

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Let f:(0,∞)→R and letF(x)=∫0x f(t)dtIf Fx2=x2(1+x) then f (4) equals

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Let $f : (0, \infty) \to R$ and let
$F (x) = \int_{0}^{x} f (t) d t$
If $F (x^{2}) = x^{2} (1 + x)$ then $f (4)$ equals