Question

Write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ x=e^{t}+2, \quad y=2 t+1 \quad-2 \leq t \leq 2 $$

Short answer

The integral that represents the arc length of the curve over the interval from -2 to 2 is \(L = \int_{-2}^{2} \sqrt{e^{2t} + 4} dt\)

Step-by-step solution

Calculate the derivatives of x(t) and y(t)

The function \(x(t) = e^{t} + 2\), differentiating it with respect to \(t\) gives \(\frac{dx}{dt} = e^{t}\). The function \(y(t) = 2t + 1\), and its derivative with respect to \(t\) is \(\frac{dy}{dt} = 2\).

Insert the derivatives into the arc length formula

Place these derivative values into the arc length formula, i.e., \(L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2}} dt\), where the interval is \(t = -2\) to \(t = 2\). So we get \(L = \int_{-2}^{2} \sqrt{(e^{t})^{2} + (2)^{2}} dt\)

Simplify the expression under the square root

Simplifying, \(L = \int_{-2}^{2} \sqrt{e^{2t} + 4} dt\) would be the integral that represents the arc length of the curve