Calc II - Definite integral of sqrt(t^2 + t) from 2x to 1?
Andrew Henderson 
How do I find
$$\int_1^{2x}\sqrt{t^2 + t}$$
with only knowledge from a Calculus I course?
I've tried plugging this puppy into Wolfram Alpha and other integral solvers, which report it as solvable (looks really long and nasty), but I think this is outside the scope of my just-entered-Calc-II knowledge and that I need to solve it in a tricky way.
The problem is part of the linked green sheet. Since I don't know how to integrate this I've tried to solve the sheet without doing so, though on part D) it seems like my luck is about to run out. I doubt what I am trying to do is even legal. Please advise.
1 Answer
$\begingroup$We have, by setting $t=2u$, $$ F(x)=\int_{1}^{2x}\sqrt{t^2+t}\,dt = 2\int_{1/2}^{x}\sqrt{4u^2+2u}\,dt$$ so:
a) $$F'(x) = 2\sqrt{4x^2+2x}$$b) The domain of $F(x)$ is $[0,+\infty)$, since $4u^2+2u$ is negative for any $u\in(-1/2,0)$.
c) $$\lim_{x\to 1/2}F(x)=0$$ since $\sqrt{4u^2+2u}$ is a continuous function in a neighbourhood of $u=1/2$.
d) $$L(\gamma)=\int_{1}^{2}\sqrt{1+F'(x)^2}\,dx = \int_{1}^{2}\sqrt{16x^2+8x+1}\,dx=\int_{1}^{2}(4x+1)\,dx=7.$$
$\endgroup$ 8
