Chapter 1: Problem 9
\(\int_{-1}^{1} \frac{4}{1+x^{2}} d x=\) (A) 0 (B) \(\pi\) (C) 2\(\pi\) (D) 2
Short Answer
Expert verified
(B) \(\pi\)
Step by step solution
01
- Recognize the standard integral
The given integral is similar to a standard integral \(\int \frac{a}{1+x^{2}} dx = a*tan^{-1}(x)\). Here, \(a = 4\). Thus, the integral can be rewritten as \(\int_{-1}^{1} 4*\frac{1}{1+x^{2}} d x\).
02
- Apply the standard integral
Applying the formula for the integral of \(\frac{1}{1+x^{2}}\), we get the integral is equal to \(4*[tan^{-1}(x)]_{-1}^{1}\)
03
- Evaluate the limits of the integral
Next, we substitute 1 and -1 into \(tan^{-1}(x)\), to get the definite integral, \(4*[tan^{-1}(1) - tan^{-1}(-1)]\). Since \(tan^{-1}(1) = \frac{\pi}{4}\) and \(tan^{-1}(-1) = -\frac{\pi}{4}\), the expression becomes \(4*[ \frac{\pi}{4} - (-\frac{\pi}{4} )]\).
04
- Final Calculation
The final step is to compute \(4*[ \frac{\pi}{4} + \frac{\pi}{4} ]\) which simplifies to \(4*\frac{\pi}{2}\). Hence the result is \(\pi\) which corresponds to option (B).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a fundamental concept in calculus, representing the accumulation of quantities over a given interval. They are used to calculate things like area under a curve, total accumulated quantity, and other similar physical concepts. A definite integral is usually written in the form \[\int_{a}^{b} f(x) \; dx\]where \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function being integrated. The result of a definite integral is a number that depends on the limits and the function itself. It's called 'definite' because it results in a specific value as opposed to an indefinite integral, which results in a function or family of functions.
- Limits of integration: The values \(a\) and \(b\) on the integral. These limits specify where the accumulation starts and ends.
- Function to be integrated: \(f(x)\), the expression we want to accumulate over \([a, b]\).
- Definite value: The result of the definite integral is a fixed number, denoting the total accumulation over the given interval.
Trigonometric Substitution
Trigonometric substitution is a great technique for simplifying integral calculations, especially when dealing with radicals or expressions involving \(1+x^2\), \(1-x^2\), or related forms. It employs trigonometric identities to transform a complex integral into a more manageable form. To accomplish this, one replaces a variable with a trigonometric function. For example, substituting \(x = \tan(\theta)\) can be useful for forms like \(1+x^2\) because it simplifies to \(1+\tan^2(\theta) = \sec^2(\theta)\). This process turns challenging algebraic expressions into simpler trigonometric ones, which can be integrated more easily.
- Useful for integrals involving square roots or quadratic forms related to trigonometric identities.
- Substitution helps by converting the original expression into a simpler trigonometric function.
- Post-integration, it’s vital to convert back using the original algebraic terms.
Inverse Trigonometric Functions
Inverse trigonometric functions are crucial in calculus, particularly when solving integrals involving expressions like \(\frac{1}{1+x^2}\). They allow us to express angles as functions of other values, essentially finding the angle whose trigonometric function equals a specified value. Common inverse trigonometric functions include \(\tan^{-1}(x)\), \(\sin^{-1}(x)\), and \(\cos^{-1}(x)\).For instance, \[\tan^{-1}(x)\]represents the angle in radians whose tangent is \(x\). These functions are typically used to solve integrals of specific forms due to their properties and relationship with differentiation.
- Useful in evaluating integrals where direct integration isn’t straightforward.
- They serve as the antiderivatives for certain rational functions.
- In our example, \(\frac{4}{1+x^2}\) led to using \(\tan^{-1}(x)\) because its derivative is \(\frac{1}{1+x^2}\).