Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the \(4 \times 4\) matrix given in the ORS. Column I (A) \(\int_{-1}^{1} \frac{d x}{1+x^{2}}\) (B) \(\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\) (C) \(\int_{2}^{3} \frac{d x}{1-x^{2}}\) (D) \(\int_{1}^{2} \frac{d x}{x \sqrt{x^{2}-1}}\) Column II (p) \(\frac{1}{2} \log \left(\frac{2}{3}\right)\) (q) \(2 \log \left(\frac{2}{3}\right)\) (r) \(\frac{\pi}{3}\) (s) \(\frac{\pi}{2}\)

Short Answer

Expert verified
Integral A matches with (s), Integral B matches with (s), Integral C matches with (p), Integral D matches with (r).

Step by step solution

01

Evaluate Integral A

Integral A can be recognized as the standard form \(\int \frac{dx}{1+x^2} = \arctan(x) + C\). Hence, evaluating at the limits from -1 to 1 we get \(\arctan(1) - \arctan(-1) = \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) = \frac{\pi}{2}\).
02

Match Integral A to Column II

The evaluated result of Integral A, \(\frac{\pi}{2}\), matches with option (s) in Column II.
03

Evaluate Integral B

Integral B can be identified as the integral for the arc length of a quarter circle with radius 1. Hence, evaluating from 0 to 1 we get \(\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}\).
04

Match Integral B to Column II

The evaluated result of Integral B, \(\frac{\pi}{2}\), also matches with option (s) in Column II.
05

Evaluate Integral C

Integral C is the inverse of the derivative of the inverse hyperbolic function \(\int \frac{dx}{1-x^2}\), which leads to \(a\,\log\left|\frac{1-x}{1+x}\right| = a\,\log\left(\frac{1-x}{1+x}\right)\) for \((x > 1)\). Evaluating from 2 to 3 gives \(\log\left(\frac{1}{5}\right) - \log\left(\frac{1}{3}\right) = \log\left(\frac{2}{3}\right)\). Hence, dividing by 2 to match the form in Column II, we get \(\frac{1}{2}\log\left(\frac{2}{3}\right)\).
06

Match Integral C to Column II

The evaluated result of Integral C, \(\frac{1}{2} \log \left(\frac{2}{3}\right)\), matches with option (p) in Column II.
07

Evaluate Integral D

For Integral D, consider the substitution \(x = \sec(\theta)\), which leads to \(dx = \sec(\theta)\tan(\theta)d\theta\). The integral transforms to \(\int \sec(\theta) \tan(\theta) / (\sec(\theta) \sqrt{\sec^2(\theta)-1}) d\theta\), which simplifies to \(\int \frac{d\theta}{\sec(\theta)}\) or \(\int \cos(\theta) d\theta\). Evaluating this integral from \(\sec^{-1}(1)\) to \(\sec^{-1}(2)\) gives us the limit from 0 to \(\cos^{-1}\left(\frac{1}{2}\right)\), which is the value of \(\frac{\pi}{3}\).
08

Match Integral D to Column II

The evaluated result of Integral D, \(\frac{\pi}{3}\), matches with option (r) in Column II.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
A definite integral is a fundamental concept in calculus that refers to the evaluation of the integral over a specific interval. It is represented by the integral sign with upper and lower limits, indicating the bounds of integration. For instance, when you come across an expression like \(\int_{a}^{b} f(x) \, dx\), it is understood as the accumulation of the function \(f(x)\) over the interval \[a, b\].

The result of a definite integral is a number that represents the area under the function's graph between the given limits. It is pivotal in various fields including physics, engineering, and economics for calculating quantities such as area, volume, and other sums involving continuous change.
Integration Techniques
Integration techniques are various methods used to solve integral problems. They are crucial for finding the antiderivatives of functions. Some common techniques include:
  • Substitution: Involves changing the variable of integration to simplify the integrand.
  • Integration by Parts: Based on the product rule for differentiation and is useful for integrating the product of two functions.
  • Partial Fractions: Used when dealing with rational functions, where the numerator’s degree is less than the denominator’s.
  • Trigonometric Substitution: Utilized when the integral includes terms that resemble Pythagorean identities.
Understanding and applying these techniques skillfully is critical for solving more complex integral problems often encountered in advanced mathematics.
Inverse Trigonometric Functions
Inverse trigonometric functions, also known as arc functions, allow us to find the angle for a given trigonometric value. They are the inverses of the basic trigonometric functions and provide solutions to trigonometric equations. For instance, \(\arcsin(x)\), \(\arccos(x)\), and \(\arctan(x)\) are inverse functions of \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\), respectively.

These functions are essential in integration because they appear as the result when we integrate expressions involving trigonometric function derivatives. For example, the integral of \(\frac{1}{1+x^2}\) with respect to \(x\) is \(\arctan(x)\) plus a constant of integration. Hence, they can also be used directly in integration to reverse-engineer the angle from its sine, cosine, or tangent value.
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. They include \(\sinh(x)\), \(\cosh(x)\), \(\tanh(x)\), and their inverses, and have properties that closely resemble trigonometric functions, making them useful in integration.

Inverse hyperbolic functions, like \(\text{arsinh}(x)\) or \(\text{arcosh}(x)\), are used when integrating expressions that have forms similar to those of the derivatives of hyperbolic functions. They appear naturally in certain integrals, much like inverse trigonometric functions, and are applied to problems dealing with rapid growth or decay, or the geometry of a hyperbola.

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