It opens downwards so it looks like this “n”. That is because a in the formula is a negative number in this situation
Answer: There are no fractions listed, so I can't help you. 1/3 is a repeating though, if that helps anything. Otherwise, you could use a calculator and just find the decimal of the fraction.
Answer:
then subtitute x to find <FGH
the awnser is -6
im very sure let me know if im wrong.

By the fundamental theorem of calculus,

Now the arc length over an arbitrary interval

is

But before we compute the integral, first we need to make sure the integrand exists over it.

is undefined if

, so we assume

and for convenience that

. Then