Question

Find the interval(s) of upward concavity on this accumulation function.

Please help and thanks in advance. Brainliest to best answer

Answer 1

$

\displaystyle

f(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\

=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\

=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\

=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\

=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\

=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\

=2x\tan(x)+\ln(\cos^2(x))

$

Hope this helps.