/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 22

Evaluate the following integrals. $$\int \frac{d x}{(x-a)(x-b)}, a \neq b$$

Short Answer

Expert verified
Based on the given step-by-step solution, provide a short answer: To find the integral of the rational function \(\frac{1}{(x-a)(x-b)}\) with \(a \neq b\), we first decompose it using partial fraction decomposition. We then find the unknown constants \(A\) and \(B\) and rewrite the function. Finally, we integrate the decomposed function and obtain the result: $$\int \frac{1}{(x-a)(x-b)} dx = \frac{\log\left|\frac{x-a}{x-b}\right|}{a-b} + C$$ where \(C\) is the integration constant.

Step by step solution

01

Perform Partial Fraction Decomposition

To decompose the given function, we can write: $$\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$ where \(A\) and \(B\) are constants. To find these constants, we will multiply the entire equation with the common denominator and simplify: $$1 = A(x-b) + B(x-a)$$ Now, let's find \(A\) and \(B\).
02

Finding the constants A and B

To find \(A\), we can set \(x = a\): $$1 = A(a-b) + 0$$ So, \(A = \frac{1}{a-b}\). To find \(B\), we can set \(x = b\): $$1 = 0 + B(b-a)$$ So, \(B = \frac{1}{b-a} = -\frac{1}{a-b}\). Now that we have the constants, we can rewrite the original function as: $$\frac{1}{(x-a)(x-b)} = \frac{1}{a-b}\left(\frac{1}{x-a} - \frac{1}{x-b}\right)$$
03

Calculating the integral

Now, let's integrate the decomposed function with respect to \(x\): $$\int \frac{1}{(x-a)(x-b)} dx = \int \frac{1}{a-b} \left(\frac{1}{x-a} - \frac{1}{x-b}\right) dx$$ $$= \frac{1}{a-b} \int \left(\frac{1}{x-a} - \frac{1}{x-b}\right) dx$$ The integral of \(\frac{1}{x-k}\) is \(\log|x-k|+C\), where \(k\) is a constant and \(C\) is the integration constant. Therefore, we have: $$\frac{1}{a-b} \left(\int\frac{1}{x-a} dx - \int\frac{1}{x-b} dx\right)$$ $$= \frac{1}{a-b}\left(\log|x-a|-\log|x-b|\right)+C$$ By using the property \(\log(a)-\log(b)=\log\frac{a}{b},\) we get the final result:
04

Final result

$$\int \frac{1}{(x-a)(x-b)} dx = \frac{\log\left|\frac{x-a}{x-b}\right|}{a-b}+ C$$

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Key Concepts

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

Partial Fraction Decomposition
Partial fraction decomposition is an essential algebraic technique used in calculus to simplify complex rational expressions for integration. This method is particularly useful when dealing with fractions that have polynomial denominators. The idea is to break down a complex fraction into a sum or difference of simpler fractions, whose denominators are the factors of the original denominators.

Through this process, integrals that seem difficult to tackle at first glance become easier to manage by transforming them into a sum of integrals with simpler integrands. For example, a given fraction \( \frac{1}{(x-a)(x-b)} \) where \( a \) and \( b \) are constants and \( a eq b \), can be expressed as a sum of two fractions with constants \( A \) and \( B \) in the numerators: \( \frac{A}{x-a} + \frac{B}{x-b} \). Determining these constants \( A \) and \( B \) involves a bit of algebra, but once they're found, integration becomes significantly straightforward.
Integration Techniques
Integration techniques involve a variety of strategies to find integrals, which are the mathematics counterpart to differentiation and are used to calculate areas, volumes, and other concepts that arise in various fields, such as physics and engineering. Effective integration can sometimes be a blend of multiple methods such as substitution, integration by parts, trigonometric integration, and of course, partial fraction decomposition which was previously discussed.

For the integral \( \int \frac{d x}{(x-a)(x-b)} \), partial fraction decomposition has converted a complex rational function into a simpler form that can be integrated using basic integral rules. Knowing that the integral of \( \frac{1}{x-k} \) is \( \log|x-k|+C \) where \( k \) is a constant, allows us to easily integrate the decomposed fraction term-by-term. This step-by-step transformation demonstrates the power of integration techniques in breaking down complex problems into manageable pieces.
Definite Integrals
Definite integrals are a core concept in calculus that represent the net area under a curve within given bounds. This concept extends beyond finding the antiderivative and involves evaluating the integral at the upper and lower limits of integration. The result of a definite integral gives us tangible information such as the displacement of an object moving along a line, when provided with its velocity-time graph, or the total accumulation of a quantity over a specified interval.

While the exercise provided does not deal with a definite integral but an indefinite one, understanding the connection is important. If given bounds, say \( [c, d] \), the final step after finding the antiderivative would be to evaluate \( \frac{\log\left|\frac{x-a}{x-b}\right|}{a-b}+ C \) at \( x = d \) and \( x = c \) and then subtract these values. This would give the exact net area between the curve \( y = \frac{1}{(x-a)(x-b)} \) and the x-axis from \( x = c \) to \( x = d \).

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Most popular questions from this chapter

An identity Show that if \(f\) and \(g\) have continuous second derivatives and \(f(0)=f(1)=g(0)=g(1)=0,\) then $$\int_{0}^{1} f^{\prime \prime}(x) g(x) d x=\int_{0}^{1} f(x) g^{\prime \prime}(x) d x.$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution \(u=\tan (x / 2)\) or, equivalently, \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1-\cos x}$$.

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty)\). a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\). b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\). c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

The gamma function is defined by \(\Gamma(p)=\int_{0}^{\infty} x^{p-1} e^{-x} d x,\) for \(p\) not equal to zero or a negative integer. a. Use the reduction formula $$\int_{0}^{\infty} x^{p} e^{-x} d x=p \int_{0}^{\infty} x^{p-1} e^{-x} d x \quad \text { for } p=1,2,3, \ldots$$ to show that \(\Gamma(p+1)=p !(p\) factorial). b. Use the substitution \(x=u^{2}\) and the fact that \(\int_{0}^{\infty} e^{-u^{2}} d u=\frac{\sqrt{\pi}}{2}\) to show that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Period of a pendulum A standard pendulum of length \(L\) that swings under the influence of gravity alone (no resistance) has a period of $$ T=\frac{4}{\omega} \int_{0}^{\pi / 2} \frac{d \varphi}{\sqrt{1-k^{2} \sin ^{2} \varphi}} $$ where \(\omega^{2}=g / L, k^{2}=\sin ^{2}\left(\theta_{0} / 2\right), g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(\theta_{0}\) is the initial angle from which the pendulum is released (in radians). Use numerical integration to approximate the period of a pendulum with \(L=1 \mathrm{m}\) that is released from an angle of \(\theta_{0}=\pi / 4\) rad.
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