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Magazine Chapter 6: Problem 31
\(\int_{-1}^{3}|x| d x=\) (A) \(\frac{7}{2}\) (B) 4 (C) \(\frac{9}{2}\) (D) 5
Short Answer
Expert verified
The integral is (D) 5.
Step by step solution
01
Understand the Absolute Value Function
The function inside the integral is \(|x|\), which has different expressions based on the value of \(x\). Specifically: \(|x| = -x\) for \(x < 0\) and \(|x| = x\) for \(x \, \geq \, 0\).
02
Break Down the Integral into Intervals
Since the limits of integration are from \(-1\) to \(3\) and the absolute value function changes at zero, we need to split the integral: 1. From \(-1\) to \(0\) (where \(|x| = -x\)) 2. From \(0\) to \(3\) (where \(|x| = x\))
03
Integrate Over the First Interval
Calculate the integral \(\int_{-1}^{0} -x \, dx\):- The antiderivative of \(-x\) is \(-\frac{x^2}{2}\).- Evaluating at \(0\) and \(-1\): \(-\left[ \frac{(0)^2}{2} \right] + \left[ \frac{(-1)^2}{2} \right] = \frac{1}{2}\).
04
Integrate Over the Second Interval
Calculate the integral \(\int_{0}^{3} x \, dx\):- The antiderivative of \(x\) is \(\frac{x^2}{2}\).- Evaluating at \(3\) and \(0\): \([\frac{(3)^2}{2}] - [\frac{(0)^2}{2}] = \frac{9}{2}\).
05
Add Results from Both Intervals
Add the results of the two integrals: \(\frac{1}{2} + \frac{9}{2} = 5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
The absolute value function, denoted as \(|x|\), outputs the distance of a number from zero, on a number line. It's a piecewise function, meaning it is defined by different expressions over different intervals. When \(x\) is negative, the value of \(|x|\) is \(-x\), because taking away the negative sign gives the distance from zero. When \(x\) is zero or positive, \(|x| = x\), as the distance remains the same.
- For all \(x < 0\), \(|x| = -x\).
- For all \(x \geq 0\), \(|x| = x\).
Definite Integral
The definite integral, denoted \( \int_{a}^{b} f(x) \, dx \), represents the net area under the curve of \(f(x)\) from \(x=a\) to \(x=b\). In essence, it sums up continuous infinitesimal products of function value and small change in \(x\).
- It evaluates to a number, unlike indefinite integrals that include a constant \(+C\).
- We find this by computing the antiderivatives, then taking the difference between the upper and lower bounds (Fundamental Theorem of Calculus).
Antiderivative
An antiderivative of a function \(f(x)\) is another function \(F(x)\) such that when you differentiate \(F(x)\), you get back \(f(x)\). The antiderivative is essential for solving integrals.
- For any constant \(C\), \(F(x) + C\) is a valid antiderivative, as differentiation of a constant is zero.
- The antiderivative helps to directly compute definite integrals.
- For \(-x\), the antiderivative is \(-\frac{x^2}{2}\).
- For \(x\), it is \( \frac{x^2}{2} \).
Piecewise Function
A piecewise function is one that has different expressions for different parts or intervals of its domain.
- It's defined using separate conditions for different sections of input values.
- This approach helps in handling situations like the absolute value function, where the form of the function changes based on input.
- On \([-1, 0]\), \(|x| = -x\).
- From \([0, 3]\), \(|x| = x\).
