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

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

Answer 2

`\bf \displaystyle \cfrac{1}{5-2}\int\limits_(2)^(5)\ \cfrac{t}{√(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}{√(u)}\cdot \cfrac{du}{2t}\implies \cfrac{1}{3}\cdot \cfrac{1}{2}\int\limits_(2)^(5)\ u^{-(1)/(2)}\cdot du\implies \cfrac{1}{6}\int\limits_(2)^(5)\ u^{-(1)/(2)}\cdot du \\\\\\ \left. \cfrac{√(u)}{3} \right]_(2)^5\implies \cfrac{√(5)}{3}-\cfrac{√(2)}{3}\implies \cfrac{√(5)-√(2)}{3}`