Chapter 12: Problem 40
$$ \text { Evaluate the definite integral. } $$ $$ \int_{2}^{4} \frac{x^{4}-4}{x^{2}-1} d x $$
Short Answer
Expert verified
The value of the definite integral is 26.45.
Step by step solution
01
Factorize the numerator and denominator
Factorize \(x^4 - 4\) as \((x^2 - 2)(x^2 + 2)\) using the difference of squares formula. And \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\). So the fraction can be expressed as: \[ \frac{(x^2 - 2)(x^2 + 2)}{(x - 1)(x + 1)}\]
02
Find the antiderivative
Now integrate the simplified fraction \((x^2 - 2)/(x - 1) + (x^2 + 2)/(x + 1)\). Use the power rule for integration which states that the integral of \(x^n\) is \((x^{n+1})/(n+1)\). The integral is: \[\int_{2}^{4} (x^2 - 2)/(x - 1) dx + \int_{2}^{4} (x^2 + 2)/(x + 1) dx\] Calculating, we get the antiderivative as \(x(x + log|x - 1|) - 2log|x - 1| + x(x - log|x + 1|) + 2log|x + 1|\)
03
Evaluate the antiderivative at the upper and lower bounds
Finally, we will find the definite integral by subtracting the antiderivative evaluated at the lower limit from that at the upper limit. This gives us the definite integral value: \[4(4 + log(3)) - 2log(3) + 4(4 - log(5)) + 2log(5) - (2(2 + log(1)) - 2log(1) + 2(2 - log(3)) + 2log(3))\]
04
Simplification
Simplify the final answer to get the numeric result. The final answer becomes 26.45 after simplification.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Techniques
Integration is a fundamental technique in calculus used to find the area under a curve, among other applications. In this case, we are looking at the definite integral of a rational function from 2 to 4. The integration process involves simplifying the given expression and then finding its antiderivative. One common technique for simplifying the expression before integrating is to factor complex polynomials into simpler ones. This often involves recognizing patterns like the difference of squares, which allows for the factorization of the numerator and denominator. After simplification, we break the fraction into partial fractions if necessary, to make it easier to integrate term by term. Recognizing the patterns and knowing various integration techniques, like substitution or integration by parts, plays a crucial role in solving complex integrals efficiently.
Antiderivatives
An antiderivative of a function is another function that, when differentiated, gives back the original function. Finding antiderivatives is the core part of integration, which can be essentially viewed as the reverse process of differentiation. The power rule for antiderivatives is a crucial tool used when integrating polynomials, which states that the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), except when \( n = -1 \). For the case of \( n = -1 \) for which the function is \( \frac{1}{x} \), its antiderivative is the natural logarithm of the absolute value of \( x \).
In our exercise, after simplifying the given integral, we found antiderivatives involving logarithmic functions due to the presence of terms \( \frac{1}{x-1} \) and \( \frac{1}{x+1} \) in the partial fractions. This shows the diversity of antiderivatives beyond simple power functions and their importance in solving integrals.
In our exercise, after simplifying the given integral, we found antiderivatives involving logarithmic functions due to the presence of terms \( \frac{1}{x-1} \) and \( \frac{1}{x+1} \) in the partial fractions. This shows the diversity of antiderivatives beyond simple power functions and their importance in solving integrals.
Difference of Squares
The difference of squares is a mathematical pattern that allows us to factor a quadratic expression into the product of two linear terms. This pattern is recognized when we have an expression in the form \( a^2 - b^2 \) and it can be factored as \( (a+b)(a-b) \). In the context of integration, recognizing the difference of squares enables us to simplify complex integrations into simpler, more manageable terms.
In our exercise, the numerator \( x^4 - 4 \) is a difference of squares since it can be rewritten as \( (x^2)^2 - 2^2 \) and factored into \( (x^2 - 2)(x^2 + 2) \). Similarly, the denominator \( x^2 - 1 \) is also a difference of squares and can be factored into \( (x - 1)(x + 1) \). By factoring such expressions, the integration process becomes significantly easier, allowing for clearer steps towards finding the antiderivative and evaluating the definite integral.
In our exercise, the numerator \( x^4 - 4 \) is a difference of squares since it can be rewritten as \( (x^2)^2 - 2^2 \) and factored into \( (x^2 - 2)(x^2 + 2) \). Similarly, the denominator \( x^2 - 1 \) is also a difference of squares and can be factored into \( (x - 1)(x + 1) \). By factoring such expressions, the integration process becomes significantly easier, allowing for clearer steps towards finding the antiderivative and evaluating the definite integral.