Problem 53 - Calculus: Early Transcendental Functions

Question

Generalize exercise 51 to \(I=\int_{0}^{d} \frac{f(x)}{f(x)+f(a-x)} d x\) for any positive, continuous function \(f\) and then quickly evaluate \(\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x\)

Step-by-Step Solution

Answer

The integral \(\int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x} dx = \frac{\pi}{4}\).

Step 1: Generalize Exercise 51

Generalizing exercise 51 we set \(y = a - x\). This will switch the limits of integration on the integral and change the differential from \(dx\) to \(-dy\). Now the integral becomes \(- \int_{a}^{0} \frac{f(a - y)}{f(a - y) + f(y)} dy\). Adding this to our original integral gives \(0 = I - I = \int_{0}^{a} \frac{f(x) - f(a - x)}{f(x) + f(a - x)} dx\). As the numerator is \(0\), we conclude that \(I = \frac{a}{2}\).

Step 2: Evaluation of the given Integral

The integral \(\int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x} dx\) can be seen as the special case of the generalized exercise 51 for \(d = \pi/2\), \(a = \pi/2\) and \(f(x) = \sin(x)\). Therefore, the integral evaluates to \(I = \frac{a}{2} = \frac{\pi}{4}\).

Key Concepts

Definite Integral

The concept of the definite integral is central to calculus and analysis, relating to the accumulation of quantities, such as areas under curves and total change. To understand a definite integral, imagine adding up an infinite number of infinitesimally thin rectangles under a curve from a starting point, denoted as the lower limit of integration, to an ending point, known as the upper limit of integration.

The notation \(\int_{a}^{b} f(x)dx\) represents the definite integral of function \(f(x)\) from \(a\) to \(b\). Calculating this requires determining the limit of a sum of products of the function values and small width intervals as these widths approach zero. If the function \(f\) is continuous on the interval \[a, b\], the Fundamental Theorem of Calculus ensures that the definite integral can be evaluated using an antiderivative of \(f\).

In simpler terms, a definite integral gives you the total accumulation of a function across an interval. It's like measuring the total water flow through a pipe from one moment to another; every drop of water is counted in the total flow.

Symmetry in Integrals

Symmetry in integrals can be a powerful tool when dealing with definite integrals of functions that exhibit reflective properties. When a function displays symmetry about the y-axis, it is even, meaning \(f(x) = f(-x)\). If a function is symmetric about the origin, it is odd, demonstrated through \(f(x) = -f(-x)\).

For even functions, you can simplify the definite integral over a symmetric interval around the origin by doubling the integral from 0 to the upper limit. For odd functions, the definite integral from \( -a \) to \( a \) is always zero, because the areas above and below the x-axis cancel each other. This property is used to simplify the calculation of integrals and can be seen in the exercise where the substitution \( y = a - x \) creates symmetry that ultimately leads to the precise value of the integral \(I\).

Understanding the symmetry of a function can save time and effort, turning complex integrals into straightforward calculations and sometimes even allowing the direct determination of an integral's value without explicit computation.

Trigonometric Integrals

Trigonometric integrals involve finding the antiderivatives of trigonometric functions. These integrals often require specific techniques and identities to simplify and evaluate them. In our exercise, the integral of \( \frac{\sin x}{\sin x + \cos x} \) is a type of trigonometric integral. Trigonometric identities like \( \sin^2(x) + \cos^2(x) = 1 \) or angle addition formulas can sometimes be used to manipulate the integrand into a more convenient form.

Another common strategy is to use symmetry, as mentioned earlier, or periodic properties of trigonometric functions to evaluate the integral over a specific interval. In some cases, substitution is a valuable technique, such as setting \( u = \sin x \) or \( u = \cos x \) and then finding the antiderivative in terms of \( u \) before converting back to the original variable.A sound strategy is to recognize patterns and apply the right trigonometric identities. These patterns often lead to significant simplifications that make the integration possible.