Question

Hello all, this is for an assignment in my Calculus 2 class. I'm submitting it online, and it keeps saying my answer is wrong :(

Answer 1

$\bf \displaystyle \cfrac{1}{5-2}\int\limits_{2}^{5}\ \cfrac{t}{\sqrt{5+t^2}}\cdot dt\\\\ -----------------------------\\\\ u=5+t^2\implies \cfrac{du}{dt}=2t\implies \cfrac{du}{2t}=dt\\\\ -----------------------------$

$\bf \displaystyle \cfrac{1}{3}\int\limits_{2}^{5}\ \cfrac{t}{\sqrt{u}}\cdot \cfrac{du}{2t}\implies \cfrac{1}{3}\cdot \cfrac{1}{2}\int\limits_{2}^{5}\ u^{-\frac{1}{2}}\cdot du\implies \cfrac{1}{6}\int\limits_{2}^{5}\ u^{-\frac{1}{2}}\cdot du \\\\\\ \left. \cfrac{\sqrt{u}}{3} \right]_{2}^5\implies \cfrac{\sqrt{5}}{3}-\cfrac{\sqrt{2}}{3}\implies \cfrac{\sqrt{5}-\sqrt{2}}{3}$

Answer 2

There IS SOMETHING WIERD  in solving the integral. Lt's solve it again:
g(t) =t / √(5+t²); solve it by substitution:; u=5+t²==> du=2tdt & t.dt = du/2

∫ [t / √(5+t²)]dt  ==> ∫(1/2 .dt / √u ==> 1/2∫(u^(-1/2) ==> 1/2[(u^1/2) / (1/2)
==> u^(1/2) ==> √u +c = √5+t²

so integral of g(t) = √(5+t²) +c.
From here & onward you can start finding the average area or value