Mathematics - Definite Integration Question with Solution | TestHub

The given integral is ${\int }_0^x\left(f^{\prime \prime }(t)+2t{f}^{\prime }(t)+t^2{f}^{\prime \prime }(t)\right)dt=k$.

Differentiating with respect to $x$:

$(1+x^2)f^{\prime \prime }(x)+2x{f}^{\prime }(x)=0$

This can be rewritten as $\frac{d}{dx}((1+x^2)f^{\prime }(x))=0$.

Integrating, $(1+x^2)f^{\prime }(x)=\lambda$.

Given $f^{\prime }(1)=1/2$, we have $(1+1^2)(1/2)=\lambda \Rightarrow \lambda =1$.

So, $f^{\prime }(x)=\frac{1}{1+x^2}$.

Integrating again, $f(x)={\tan }^{-1}(x)+C$.

Given $f(0)=0$, we have ${\tan }^{-1}(0)+C=0\Rightarrow C=0$.

Thus, $f(x)={\tan }^{-1}(x)$.

We need to find ${\lim }_{x\to \pi /4^{+}}[f(x)]={\lim }_{x\to \pi /4^{+}}[{\tan }^{-1}(x)]$.

Since $\pi /4\approx 0.785$, and ${\tan }^{-1}(x)$ is an increasing function, for $x>\pi /4$ but close to $\pi /4$, ${\tan }^{-1}(x)>{\tan }^{-1}(\pi /4)$.

${\tan }^{-1}(\pi /4)\approx {\tan }^{-1}(0.785)\approx 0.665$.

So, for $x\to \pi /4^{+}$, $f(x)$ will be slightly greater than $0.665$.

Therefore, $[f(x)]=[\mathrm{0.665...}]=0$.

The final answer is $\boxed{0}$.