If f(x) is a differentiable function defined for all positive real numbers such that x⋅f(x)=x+∫1xf(t)dt , then the value of ∑k=110fek is___

x⋅f'(x)+f(x)=1+f(x)⇒f'(x)=1xf(x)=lnx+k But f(1)=1∴k=1∴f(x)=lnx+1 k=110f(ek)= k=110lnek+1= k=110(k+1)=65

If f(x) is a differentiable function defined for all positive real numbers such that x⋅f(x)=x+∫1xf(t)dt , then the value of ∑k=110fek is___

x⋅f'(x)+f(x)=1+f(x)⇒f'(x)=1xf(x)=lnx+k But f(1)=1∴k=1∴f(x)=lnx+1 k=110f(ek)= k=110lnek+1= k=110(k+1)=65

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If $f (x)$ is a differentiable function defined for all positive real numbers such that $x \cdot f (x) = x + {\int^{}}_{1}^{x} f (t) d t$ , then the value of $\sum_{k = 1}^{10} f (e^{k})$ is___

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Detailed Solution

$x \cdot f^{'} (x) + f (x) = 1 + f (x) \Rightarrow f^{'} (x) = \frac{1}{x}$

$f (x) = \ln x + k$

$B u t f (1) = 1 ∴ k = 1$

$∴ f (x) = \ln x + 1$

$_{k = 1}^{10} f (e^{k}) =_{k = 1}^{10} \ln e^{k} + 1$

$=_{k = 1}^{10} (k + 1) = 65$

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