Christmas Streak 83/88: Calculus Rush #1

Note that $f(x)$ is symmetrical to $x=1.$

Then we can see that $\displaystyle \int_{0}^{2}f(x)dx=10~\Leftrightarrow~g(2)=10.$

Since $g(x)$ is an indefinite integral of $f(x),$ we get that $g(x)$ is symmetrical to a point $(1,~g(1))=(1,~5).$

Therefore we conclude that $\displaystyle \int_{0}^{2}g(x)=10.$

Now look at the thing we're trying to find the value of.

$\begin{aligned} &\int_{0}^{4}g'(\sqrt{x})dx ~\Rightarrow~x=t^2 \\ \\ &=\int_{0}^{2}2tg'(t)dt \\ \\ &=2\left(\left[tg(t)\right]_0^2 - \int_{0}^{2}g(t)dt\right) \\ \\ &=2(2g(2)-10) \\ \\ &=\boxed{20}. \end{aligned}$