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Magazine Chapter 5: Problem 51
Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{0}^{2} \frac{x^{2}-2}{x+1} d x $$
Short Answer
Expert verified
The result of the integral is \( -\ln3 \).
Step by step solution
01
Polynomial Long Division
The integral is of a rational function in the form \( \frac {f(x)}{g(x)} \). When the degree of the numerator, \( f(x) \), is equal to or greater than the degree of the denominator, \( g(x) \), we usually perform a polynomial long division. Divide the numerator by the denominator, thus: \( \frac{x^{2}-2}{x+1} = x-1+ \frac{-x-1}{x+1} \).
02
Simplify the Integral
Rewrite the integral from Step 1 as the sum of two simpler parts as: \( \int_{0}^{2} (x-1) dx + \int_{0}^{2} \frac{-x-1}{x+1} dx \). It's easier to find the integral of each of these simpler functions.
03
Integrate the First Part
Evaluate the integral \( \int_{0}^{2} (x-1) dx \) first. The antiderivative of \( x-1 \) is \( \frac{x^{2}}{2}-x \). By the Fundamental Theorem of Calculus, the definite integral from 0 to 2 is given by: \( [\frac{2^{2}}{2} - 2] - [\frac{0^{2}}{2}-0] =2-2=0 \).
04
Integrate the Second Part
Evaluate the integral \( \int_{0}^{2} \frac{-x-1}{x+1} dx \). The antiderivative of \( \frac{-x-1}{x+1} \) requires the rule for integrating \( u^{-1} \) which gives \( \ln|u| \), where here \( u = x+1 \). Therefore, the integral is \( -\ln|x+1| \) and by the Fundamental Theorem of Calculus, its value from 0 to 2 is: \( -[\ln|3|] - [ -\ln|1|] =-\ln3 \).
05
Sum the Parts
Finally, sum the two results from Step 3 and Step 4. The result of the integral is given by: \( 0 - \ln3 = -\ln3 \).
06
Verify the Result
You can use a graphing calculator or online tool to calculate the definite integral from Step 1 to confirm if it is indeed equal to \( -\ln3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Long Division
Polynomial long division is a technique used when dividing polynomials, much like long division in arithmetic, except here, we deal with variables. In our problem, we have \( \frac{x^2 - 2}{x + 1} \). Since the degree of the numerator \( x^2 \) is higher than the degree of the denominator \( x + 1 \), we must perform polynomial long division. This method allows us to simplify the fraction into a polynomial and a remainder that is easier to integrate. Here鈥檚 a step-by-step approach:
- Divide the leading term of the numerator by the leading term of the denominator: \( \frac{x^2}{x} = x \).
- Multiply the whole divisor \( x + 1 \) by this result \( x \), getting \( x^2 + x \) and subtract from the original \( x^2 - 2 \), resulting in \( -x - 2 \).
- Repeat the process for the new polynomial \( -x - 2 \) divided by \( x + 1 \), getting \( -1 \).
- Thus, we have \( \frac{x^2 - 2}{x + 1} = x - 1 + \frac{-x - 1}{x + 1} \).
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation and integration, showing the relationship between the two processes. It states that if a function is continuous on a closed interval \([a, b]\) and \( F \) is an antiderivative of \( f \) on the interval, then the definite integral is calculated as:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]In our exercise, after simplifying \( \frac{x^2 - 2}{x + 1} \) through polynomial long division, we integrate the simpler parts. For example, the integral \( \int_{0}^{2} (x-1) \, dx \) uses this theorem, where:
- The antiderivative of \( x - 1 \) is \( \frac{x^2}{2} - x \).
- Applying the theorem gives \( \left[ \frac{2^2}{2} - 2 \right] - \left[ \frac{0^2}{2} - 0 \right] = 0 \).
Antiderivative
An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. Finding an antiderivative is essential in solving definite integrals.For instance, in the exercise, to solve \( \int_{0}^{2} (x-1) \, dx \), we find an antiderivative of \( x-1 \), which is \( \frac{x^2}{2} - x \). This function, when differentiated, returns \( x-1 \). Key points about antiderivatives:
- Every continuous function has infinitely many antiderivatives, differing by a constant.
- The antiderivative helps in determining the area under the curve between two points.
- The process involves reversing differentiation鈥攁dding up a rate of change.
Logarithmic Integration
Logarithmic integration is a technique used when integrating functions of the form \( \frac{1}{u} \). This arises in problems where substitution leads to a logarithmic function.In the second integral portion of our exercise, \( \int_{0}^{2} \frac{-x-1}{x+1} \, dx \), we noticed a form that fits the logarithmic rule. Here, substituting \( u = x + 1 \) turns the integral into a logarithmic one. Let's look at how this works:
- Set \( u = x + 1 \) leading to \( du = dx \), transforming \( \frac{-x-1}{x+1} \) to a derivative form suitable for logarithmic integration.
- The integral transforms into \( -\ln|x+1| \).
- The definite integral from 0 to 2 then becomes \(- \left[ \ln|3| - \ln|1| \right] = -\ln3 \).
