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Chapter 7: Problem 41

Use a table of integrals to solve the following problems. Find the length of the curve \(y=e^{x}\) on the interval \([0, \ln 2].\)

Short Answer

Expert verified
Answer: The length of the curve is \(L = \log(2 + \sqrt{5}) - \log(1 + \sqrt{2})\).

Step by step solution

01

Find the derivative of the function

Differentiate the function \(y = e^x\) with respect to \(x\) to find its derivative: $$ \frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x $$
02

Set up the arc length formula

The arc length formula for a curve \(y = f(x)\) on the interval \([a,b]\) is given by: $$ L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx $$ We have found the derivative of the function, which is \(\frac{dy}{dx} = e^x\). Now, substitute the given derivative and the given interval into the formula: $$ L = \int_0^{\ln 2} \sqrt{1 + (e^x)^2} dx $$
03

Simplify the integrand

Simplify the expression inside the square root: $$ L = \int_0^{\ln 2} \sqrt{1 + e^{2x}} dx $$
04

Use the substitution method

Let \(u = e^x\). Then, \(\frac{du}{dx} = e^x\) and \(dx = \frac{du}{u}\). Now, we need to change the limits of integration according to the substitution. When \(x = 0,\ u = e^0 = 1\) and when \(x = \ln 2,\ u = e^{\ln 2} = 2\). The integral becomes: $$ L = \int_1^2 \sqrt{1 + u^2} \frac{du}{u} $$
05

Check if the integral is in the table

Looking at the integral, we find that it matches the formula for arc length of a hyperbolic function: $$ \int \frac{\sqrt{1 + u^2}}{u} du = \log(u + \sqrt{1 + u^2}) $$
06

Evaluate the integral

Replace the integral with the corresponding result from the table of integrals and evaluate it in the given interval: $$ L = \log(2 + \sqrt{1 + 2^2}) - \log(1 + \sqrt{1 + 1^2}) $$
07

Simplify the final result

Simplify the expression and obtain the length of the curve: $$ L = \log(2 + \sqrt{5}) - \log(1 + \sqrt{2}) $$ The length of the curve \(y = e^x\) on the interval \([0, \ln 2]\) is \(L = \log(2 + \sqrt{5}) - \log(1 + \sqrt{2})\).

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Most popular questions from this chapter

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