91Ó°ÊÓ

Integration by parts is a powerful technique in calculus used for integrating products of functions. It is based on the product rule for differentiation and transforms the integral of a product into simpler integrals. The integration by parts formula is given by:\[\int u \, dv = uv - \int v \, du\]where:- \( u \) and \( v \) are differentiable functions of \( x \).- \( du \) and \( dv \) are their respective derivatives.

• Choose \( u \) as the function that becomes simpler when differentiated.
• Choose \( dv \) as the function that can be easily integrated.

In the provided solution, integration by parts was employed to calculate the moment about the \( y \)-axis, \(M_x\). The task was to solve the integral \(\int_{0}^{1} x \cdot \frac{1}{\sqrt{1+x^{2}}} \, dx\). Here:- Let \( u = x \), resulting in \( du = dx \).- Let \( dv = \frac{1}{\sqrt{1+x^{2}}} \, dx \), resulting in \( v \) being the antiderivative of \( dv \).This method helps break down complex integrals into manageable pieces, which is especially valuable in physics and engineering problems.

Trigonometric substitution is a method used to simplify integrals involving expressions like \(\sqrt{a^2 + x^2}\), \(\sqrt{a^2 - x^2}\), or \(\sqrt{x^2 - a^2}\). By substituting a trigonometric function for \( x \), these expressions can become trigonometric identities, which are easier to work with.To solve the integral \(\int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \, dx\), a trigonometric substitution like \(x = \tan(\theta)\) is used:

• \( dx = \sec^2(\theta) \, d\theta \)
• \(\sqrt{1+x^2} = \sec(\theta)\)

This transforms the integral into an expression in terms of \(\theta\), allowing for easier evaluation.In the original problem, this specific substitution streamlines the integration process leading directly to the result \(\sinh^{-1}(1) = \ln(\sqrt{2} + 1)\). This method is particularly useful when dealing with curves that form a right triangle, where the hypotenuse is a function involving \(x\). The trigonometric functions \(\sin\), \(\cos\), and \(\tan\) can elegantly simplify the evaluation of such integrals.

The center of mass coordinates, \((\bar{x}, \bar{y})\), represent the "average" position of mass in a two-dimensional object or area. To determine these coordinates for a planar region, we use moments about the axes and the total area under the curve.The coordinates are calculated using:

• \(\bar{x} = \frac{M_y}{A}\)
• \(\bar{y} = \frac{M_x}{A}\)

where:- \( M_x \) is the moment about the \( y \)-axis, capturing how "spread out" the area is horizontally.- \( M_y \) is the moment about the \( x \)-axis, capturing how "spread out" the area is vertically.- \( A \) is the total area of the region.By integrating across the region's boundaries, these moments effectively weigh each point based on its distance from the axis in question. In the exercise, the center of mass coordinates for the region under \(f(x)\) were calculated, ensuring that both the area and moment values contribute to finding this center balance point. Such calculations are pivotal in various fields like physics and engineering for analyzing balance, stability, and structure design.