For the function f(x) = 1 4 ex + e−x, prove that the arc length on any interval has the same value as the area under the curve. f(x) = 1 4 ex + e−x ⇒ f '(x) = ⇒ 1 + [f '(x)]2 = 1 + = 2 = [f(x)]2 The arc length of the curve y = f(x) on the interval [a, b] is L = b a 1 + [f '(x)]2 dx = b a [f(x)]2 dx = b a f(x) dx, which is the area under the curve y = f(x) on the interval [a, b].

Think of it as five or above give it a shove. change the 5 or above to the right to a zero and add one to the number on the left.

First step is to get rid of the square root. remember that a square root is the same as raising to the (1/2) power. see below: multiply exponents per exponent rules to get: remember you subtract exponents from the denominator and add exponents in the numerator so: simplify further: so a is the correct choice. if you don't understand this google exponent rules and it will you understand. look at the image results.