91Ó°ÊÓ

The concept of area under a curve is fundamental in calculus. When visualizing functions on a graph, the area under the curve indicates the accumulation of quantities, like distance traveled over time or the total response in a biological assay.
In this exercise, our task is to calculate the area enclosed by the complex curve expressed by the function \(y=\frac{1}{\sqrt{1-9x^2}}\), which behaves like a segment of an upper semi-circle. This function is plotted along with the lines \(x=0\), \(x=\frac{1}{6}\), and the x-axis, \(y=0\), forming a specific region to analyze.

• The area represents the integral of the function between specified bounds, which informs us about the total 'size' of the curve in that region.
• In our case, evaluating this involves setting up and computing a specific integral bound by the limits of integration \(x=0\) to \(x=\frac{1}{6}\).

The area provides a quantitative measure of the space enclosed, making it critical in various scientific and engineering fields.

Inverse trigonometric functions, like \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\), provide solutions to equations where we want to determine angles given a certain ratio of sides. These functions are essential when transforming differential equations involving trigonometric expressions into solveable arithmetical equations.
In our current problem, the curve \(y=\frac{1}{\sqrt{1-9x^2}}\) corresponds to the structure of an inverse sine function. Recognizing this pattern is key for simplifying the integral. Specifically:

• This curve is akin to the standard form for inverse trigonometric function integrals: \(\int \frac{1}{\sqrt{a^2-x^2}}\, dx\), simplifying to \(\sin^{-1}\left(\frac{x}{a}\right) + C\).
• Here, \(a\) becomes \(\frac{1}{3}\), allowing us to convert our problem into evaluating \(\sin^{-1}(3x)\).

Transforming complex expressions into inverse trigonometric functions greatly reduces calculation complexity, essential for efficiency in advanced mathematics.

Definite integrals are a cornerstone concept in calculus, useful for determining the exact area under a curve between two given points. Unlike indefinite integrals, which provide a family of functions representing original equations, definite integrals result in a specific value.
In this exercise, the integral \(\int_{0}^{\frac{1}{6}} \frac{1}{\sqrt{1-9x^2}}\, dx\) computes the enclosed area under our complex curve from \(x=0\) to \(x=\frac{1}{6}\).

• By evaluating this integral, we apply the antiderivative developed from the inverse trigonometric form: \(\sin^{-1}(3x)\).
• Substituting our limits, we find: \(\sin^{-1}(3 \times \frac{1}{6}) - \sin^{-1}(3 \times 0)\), simplifying to \(\frac{\pi}{6}\).

This definite integral calculation provides the precise measure of space between the curve and the x-axis, highlighting its utility in precise area determination.