Integrating $\int_{x_1}^{x_2}\sqrt{1+(2at-a(x_1+x_2))^2}\,dt$

Question

How to integrate:

$$\int_{x_1}^{x_2}\sqrt{1+(2at-a(x_1+x_2))^2}\,dt$$

I know I can use u-substitution by setting $u=1+(2at-a(x_1+x_2))^2$, or something along those lines, and then solving the simplified integral and then moving back to the "x world". I'm generally familiar with first year calculus, and maybe second year, I'm not really sure as I have offically only taken pre-calc so if you could simplify your answers a bit that would be appreciated!

Answer 1

Hint:

Note that we have: $$I =\int_{x_1}^{x_2} \sqrt{1+(2at-a(x_1+x_2))^2}\, dt$$ $$=\int_{x_1}^{x_2} \sqrt{1+a^2(2t-x_1-x_2)^2}\, dt$$

Substituting, $u=2t-x_1-x_2 \implies du = 2dt$, we have, $$I =\frac12 \int_{x_1-x_2}^{x_2-x_1} \sqrt{1+a^2u^2}\, du$$ Now, substituting $au= y\implies adu = dy$, we have, $$I = \frac{1}{2a} \int_{a(x_1-x_2)}^{a(x_2-x_1)} \sqrt{1+y^2}\, dy$$

Now, the integrand is in a standard form. Can you take it from here?

Answer 2

Take the parabola $$f(x) = at^2 - at(x_1+x_2)$$ and note that

$$\int_{x_1}^{x_2}\sqrt{1+(2at-a(x_1+x_2))^2}dt = \int_{x_1}^{x_2}\sqrt{1+(f'(t))^2}dt $$

But $ \int_{x_1}^{x_2}\sqrt{1+(f'(t))^2}dt $ represents the arc lenght of the parabola $f$ from $x_1$ to $x_2$.

Now, we also have that $f(x) = 0$ only if $x=0$ or $x=x_1+x_2$ and $f'(x) = 0$ only if $x=\dfrac{x_1+x_2}{2}$, so this integral actually represents twice the arc lenght of the parabola $f$ from $x_1$ to $(x_1+x_2)/2$.

I don't know if this information helps you, but I think it's cool this point of view.