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Chapter 7: Problem 55

$$ \begin{array}{l}{\text { Use a graph of } f(x)=1 /\left(x^{2}-2 x-3\right) \text { to decide whether }} \\ {\int_{0}^{2} f(x) d x \text { is positive or negative. Use the graph to give a }} \\ {\text { rough estimate of the value of the integral and then use partial }} \\ {\text { fractions to find the exact value. }}\end{array} $$

Short Answer

Expert verified
The integral \(\int_{0}^{2} f(x) \, dx\) is negative, exact value is \(-\frac{1}{2}\ln 3\).

Step by step solution

01

Analyze the Rational Function

The function is \( f(x) = \frac{1}{x^2 - 2x - 3} \). Start by finding the roots of the denominator to identify vertical asymptotes; solve \(x^2 - 2x - 3 = 0\). Factor the quadratic to get \((x - 3)(x + 1) = 0\), thus, the roots are \(x = 3\) and \(x = -1\). The vertical asymptotes are at these points.
02

Graph the Function

Sketch the graph of \(f(x)\). Since it has vertical asymptotes at \(x = -1\) and \(x = 3\), the behavior near these points diverges. The sign of \(f(x)\) between \(x = 0\) and \(x = 2\) can be determined by plugging in a value between these points, such as \(x = 1\). Since the polynomial in the denominator \(1/(1-2-3) = -1/4\), \(f(x)\) is negative on the interval.
03

Estimate the Integral

Discerning from the graph, the integral over \([0, 2]\) reflects an area below the x-axis due to \(f(x)\) being negative on this interval. This suggests the integral is negative. Graphically estimate it by considering the shape; observing that the curve is smooth and symmetric, the rough estimate of the area under the curve is a small negative value.
04

Partial Fraction Decomposition

Express \( f(x) \) using partial fractions: \( \frac{1}{x^2 - 2x - 3} = \frac{1}{(x-3)(x+1)} \). Decompose it into \( \frac{A}{x-3} + \frac{B}{x+1} \). Solve for A and B by clearing fractions and equating coefficients: \(1 = A(x+1) + B(x-3)\). Substitute values \(x = 3\) and \(x = -1\) to find \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\).
05

Solve the Integral

Now, integrate \( \int_0^2 \left( \frac{1}{4(x-3)} - \frac{1}{4(x+1)} \right) dx \). This yields: \[ \frac{1}{4} \left[ \ln |x-3| - \ln |x+1| \right]_0^2 \] Evaluate this to find the exact value. Substituting end points: \[ \frac{1}{4} \left( \ln|2-3| - \ln|2+1| \right) - \frac{1}{4} \left( \ln|0-3| - \ln|0+1| \right) \], compute to find \[- \frac{1}{2} \ln 3 \].
06

Compare Graphical and Exact Values

The exact value computed (-\(\frac{1}{2}\ln 3\)) is consistent with the graphical estimation that the integral is negative. This confirms both the estimated and exact evaluative approaches align with the initial graphical assessments.

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Key Concepts

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

Graphical Estimation
Graphical estimation in calculus involves using a visual representation to predict the behavior or approximate the value of a mathematical function over a set interval. In this context, students are given the task of estimating the integral of a function, like \[ f(x) = \frac{1}{x^2 - 2x - 3} \] over the interval \[ [0, 2] \].To start, sketching the graph is essential. Vertical asymptotes help identify the points where the function tends to infinity. Here, they occur at \[ x = -1 \] and \[ x = 3 \]. Between \[ x = 0 \] and \[ x = 2 \], you should determine the sign of the function, which is achieved by substituting a sample point, such as \[ x = 1 \].
  • This graphical approach reveals that the function is negative on \[ [0, 2] \], which means the area under the curve and above the x-axis also lies below the x-axis, yielding a negative integral.
  • By estimating graphically, you can predict that the value of the integral is small and negative, aligning with the behavior of the curve.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. It is particularly useful in integral calculus for simplifying the integration process by turning a single compound fraction into multiple simpler terms.Given the function \[ f(x) = \frac{1}{x^2 - 2x - 3} \], partial fraction decomposition allows us to express it as:
  • Break the denominator into factors: \[ (x-3)(x+1) \].
  • Write the expression in the form of \[ \frac{A}{x-3} + \frac{B}{x+1} \].
Solving for \( A \) and \( B \) requires clearing the fractions and equating coefficients:
  • By substituting \[ x = 3 \] and \[ x = -1 \], solve for \( A = \frac{1}{4} \) and \( B = -\frac{1}{4} \).
  • This transformation simplifies the integral, making it easier to find an exact solution.
Rational Functions
Rational functions are quotients of two polynomials, where the numerator and denominator are both polynomial expressions. The integral calculus often deals with these functions to understand their properties and behaviors.The function \[ f(x) = \frac{1}{x^2 - 2x - 3} \] is a rational function. Analyzing it began with recognizing its vertical asymptotes, which occur where the denominator is zero:
  • Solving \[ x^2 - 2x - 3 = 0 \], results in factors \[ (x-3)(x+1) = 0 \] with roots \[ x = 3 \] and \[ x = -1 \].
  • This shows the values that make the function undefined, thus impacting the function graph with vertical asymptotes at these points.
  • When considering intervals for integration, any roots or asymptotes outside the bounds do not affect computation directly but play a crucial role in understanding the graph's shape.

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