Problem 22 - A Graphical Approach to Precalculus with Limits

Question

Find the exact value of each integral, using formulas from geometry. Do not use a calculator. $$\int_{-3}^{2}|x+1| d x$$

Step-by-Step Solution

Answer

The exact value of the integral is \(-1.5\).

Step 1: Determine the points where the absolute function changes

The absolute function \(|x+1|\) equals zero when \(x+1=0\). Therefore, solve \(x+1=0\) to find the point where the function changes signs: \(x = -1\). This point divides the integral range \([-3, 2]\) into two intervals: \([-3, -1]\) and \([-1, 2]\).

Step 2: Analyze the function on each interval

For \(x < -1\), the function \(|x+1| = -(x+1) = -x-1\) since \(x+1\) is negative. For \(x \geq -1\), \(|x+1| = x+1\) because \(x+1\) is non-negative.

Step 3: Compute the integral over the first interval

The integral over the interval \([-3, -1]\) is \(-\int_{-3}^{-1}(x+1)\,dx = -\left[\frac{x^2}{2} + x\right]_{-3}^{-1}\). Calculate the definite integral: \(-\left(\left(\left(-\frac{1^2}{2} + (-1)\right)\right) - \left(\left(-\frac{9}{2} + (-3)\right)\right)\right)\).

Step 4: Simplify the first integral result

Compute \(-\left(\left(-\frac{1}{2} - 1\right) - \left(-\frac{9}{2} - 3\right)\right) = -\left(-\frac{3}{2} + \frac{15}{2}\right) = -\frac{12}{2} = -6\).

Step 5: Compute the integral over the second interval

For the interval \([-1, 2]\), integrate \(|x+1| = x+1\) to get \(\int_{-1}^{2} (x+1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{2}\). Evaluate: \(\left(\frac{4}{2} + 2\right) - \left(\frac{1}{2} - 1\right)\).

Step 6: Simplify the second integral result

Compute the result: \(\left(2 + 2\right) - \left(\frac{1}{2} - 1\right) = 4 + 0.5 = 4.5\).

Step 7: Combine the results

Add the results of the two integrals: \(-6 + 4.5 = -1.5\). Thus, the exact value of the integral \(\int_{-3}^{2}|x+1| \, dx\) is \(-1.5\).

Key Concepts

Absolute Value Function

The absolute value function is expressed as \(|x|\), and it represents the distance of a number \(x\) from zero on the number line. It's a piecewise function because it behaves differently based on whether \(x\) is positive or negative.
For the expression \(|x+1|\), think of it as measuring how far \(x+1\) is from zero. Thus, \(|x+1| = x + 1\) when \(x+1\) is non-negative (or \(x \geq -1\)), and \(|x+1| = -(x+1)\) when \(x+1\) is negative (or \(x < -1\)).

• In the interval \([-3, -1)\): \(|x+1|\) converts to \(-(x+1)\), equation becomes \(-x-1\).
• In the interval \([-1, 2]\): \(|x+1|\) turns into \(x+1\).

Understanding how \(|x+1|\) distributes itself into linear functions in different intervals is critical for correctly evaluating integrals involving absolute value functions.

Piecewise Functions

Piecewise functions are fascinating as they help in defining functions that behave differently over various intervals. In simple terms, they allow a single function to exhibit different characteristics based on the given part of its domain.
For instance, with the integral \(\int_{-3}^{2}|x+1| \, dx\), the absolute value introduces a piecewise function which changes at \(x = -1\). This point is the critical boundary where the expression \(x+1\) alternates between negative and non-negative.

• For \(x < -1\), the piece \(-x-1\) results from the negative sign within \(|x+1|\), and this governs the function.
• For \(x \geq -1\), the piece \(x+1\) is derived directly because \(x+1\) is non-negative.

Engineering functions like this are handy for analyzing more complicated systems where behavior varies with different inputs.

Area Under a Curve

The concept of finding the area under a curve becomes practical when dealing with definite integrals. The integral of a function over an interval provides the area under the curve defined by the function and the limits of integration.
In this exercise, we determine the area represented by the integral \(\int_{-3}^{2}|x+1| \, dx\). The absolute value function \(|x+1|\) splits the domain into two separate curves, each with its respective definite integral to be evaluated over the specified intervals.

• The interval \([-3, -1]\) with function \(-x-1\) gives us an area of \(-6\).
• The interval \([-1, 2]\) with function \(x+1\) results in an area of \(4.5\).

By adding these two areas, we find the total area under the curve across the entire interval, resulting in a value of \(-1.5\). This geometric approach to definite integrals delivers crucial insights into calculating areas bounded by curves over specified domains.