if $f;R\rightarrow R$ is continuous, and $f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt$, determine which of the followng is right $f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt$
and the given options are

A) $5<f(\sqrt2)<6$


b) $2<f(\sqrt2)<3$


c) $3<f(\sqrt2)<4$


d) $4<f(\sqrt2)<5$

now,as it's only continuous it's surely integrable and not necessarily diffferentiable so, we have.
$f(x)=e^{\frac{x^2}{2}}+\int_0^x tf(t)dt----1$
$e^{\frac{x^2}{2}}+t \int_0^xf(t)dt-\int_0^x\int_0^xf(t)dt$
but this doesn't help at all
I'm tempeted to differentiate to maybbe obtain a useful expression, so
$$f'(x)=2xe^{\frac{x^2}{2}}+xf(x)$$
which allows me to use integration by parts in $1$
so $f(x)=e^{\frac{x^2}{2}}+\frac{x^2f(x)}{2}-\int_0^x \frac{x^2}{2}f'(x)dx$
which can further be simplified using $f'(x)$
which gives us
$f(x)=e^{\frac{x^2}{2}}+\frac{x^2f(x)}{2}-\int_0^x \frac{x^2}{2}2xe^{\frac{x^2}{2}}+xf(x)dx$
beyond which I'm lost, as the integral seems to become zero
I'd really appreciate a HINT NOT AN EXPLICIT SOLUTON FOR NOW

Hint: You already have
$$f'(x)=2xe^{\frac{x^2}{2}}+xf(x)$$
and hence
$$ f'(x)-xf(x)=2xe^{\frac{x^2}{2}}. \tag1$$
The integral factor of (1) is
$$ \mu(x)=\exp\bigg(\int(-x)dx\bigg)=e^{-\frac{x^2}{2}} $$
and multiplying (1) by $\mu(x)$ gives
$$ \bigg(e^{-\frac{x^2}{2}}f(x)\bigg)'=2x. $$
You can do the rest.
