91Ó°ÊÓ

Integration is a core concept in calculus, and when we integrate with respect to \(x\), we are essentially finding the accumulated area under a curve as \(x\) changes. In the context of finding arc lengths, we use integration to determine the length of a curve between two points.
The formula for arc length \(L\) of a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is:

• \(L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\)

This formula is derived from the Pythagorean theorem, considering an infinitesimally small segment of the curve as a straight line triangle. By summing up these tiny segments from \(x=a\) to \(x=b\), using integration, we get the total arc length.
In the problem, we integrate from \(-\ln 2\) to \(\ln 2\) using the derivative \(y'(x)=\frac{1}{2}(e^x - e^{-x})\), plugged into the arc length formula. This gives us a full representation of the curve's length over the specified interval.

Calculus is a branch of mathematics that deals with change and motion. It consists mainly of two components: differentiation and integration. In this exercise, both of these components play a crucial role.
Differentiation is used to find the derivative \(f'(x)\) of a function, which provides the slope of the tangent to the curve at any point \(x\). This slope is essential when calculating the arc length, as it forms part of the arc length integrand formula.
Integration, on the other hand, is used to sum up an infinite number of infinitesimally small quantities, such as the arc length segments mentioned earlier. It allows us to accumulate these small lengths across the interval \([-\ln 2, \ln 2]\) to calculate the total length of the curve. Calculus bridges the gap between algebraic concepts and the geometry of graphs, enabling us to solve complex problems involving curves and rates of change.

Calculating the derivative of exponential functions is a fundamental skill in calculus, and it's essential for solving the given exercise.
An exponential function is generally of the form \(e^x\), where \(e\) is the mathematical constant approximately equal to 2.718. The unique property of the exponential function \(e^x\) is that its derivative is itself, i.e., \(\frac{d}{dx} e^x = e^x\).

• For the negative exponential, \(\frac{d}{dx} e^{-x} = -e^{-x}\), due to the chain rule of differentiation.

In the exercise, the function is \(y(x) = \frac{1}{2}(e^x + e^{-x})\). The derivative is computed term by term:

• The derivative of the first term \(\frac{1}{2} e^x\) is \(\frac{1}{2} e^x\).
• The derivative of the second term \(\frac{1}{2} e^{-x}\) is \(-\frac{1}{2} e^{-x}\).

Thus, combining these gives \(y'(x) = \frac{1}{2}(e^x - e^{-x})\), which is then used to compute the arc length. Understanding these derivatives helps in manipulating exponential functions in calculus and solving related problems efficiently.