Need help with a certain integral

Question

Sorry, this is my first post ever and formatting is bad. I appreciate all assistance

Given the equation y= $ 3.8x^2-4.4x+1444 $ Find the definite arc length integral between 0 and 1200.

I am not sure on how to format $(\frac {dy}{dx})^2$ ,so I left it alone.

$$\int_0^{1200} \sqrt{1+ (\frac{dy}{dx}})^2 dx$$

Answer 1

In the same spirit as David G. Stork'answer, consider the general case of $$y=ax^2+bx+c$$ $$\frac{dy}{dx}=2ax+b$$ So $$L=\int \sqrt{1+ (\frac{dy}{dx})^2}\, dx=\int \sqrt{1+ (2ax+b)^2}\, dx$$ Now, make a change of variable $$2ax+b=t \implies x=\frac{t-b}{2 a}\implies dx=\frac{dt}{2 a}$$ So $$L=\frac{1}{2 a}\int \sqrt{1+t^2}\,dt$$ Again, a standard change of variable $$t=\sinh(z)\implies dt=\cosh(z)\,dz$$ $$L=\frac{1}{2 a}\int \cosh^2(z)\,dz=\frac{1}{2 a}\int \frac 12 \left(1+\cosh(2z)\right)\,dz=\frac{1}{4 a}\left(z+\frac 12 \sinh(2z)\right)$$ Back to $x$, this gives $$L=\frac{(2 a x+b)\sqrt{1+(2 a x+b)^2} +\sinh ^{-1}(2 a x+b)}{4 a}$$

Answer 2

Take the derivative and plug into your integral and use the fact that

$\int \sqrt{1 + (a + b x)^2} dx = \frac{\sqrt{(a+b x)^2+1} (a+b x)+\sinh ^{-1}(a+b x)}{2 b}$.