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Chapter 5: Problem 86

Area by geometry Use geometry to evaluate the following integrals. $$\int_{1}^{6}|2 x-4| d x$$

Short Answer

Expert verified
What are the steps to evaluate the definite integral of the absolute value function: Step 1: Find the interval where the integrand is positive and negative. Step 2: Identify the functions for each interval. Step 3: Calculate the areas under the curves for each interval. Step 4: Add the areas and evaluate the integral. Using these steps, the definite integral $\int_{1}^{6}|2x - 4| dx$ is evaluated as: $$\int_{1}^{6}|2x - 4| dx = \text{Area from 1 to 2} + \text{Area from 2 to 6} = 0 + \frac{1}{2} \times 4 \times 4 = 8$$

Step by step solution

01

Find the interval where the integrand is positive and negative.

We already know the bounds of the integral are between 1 and 6. To determine when the integrand, |2x - 4|, is positive or negative, we can check for the point where the function inside the absolute value becomes zero: $$2x - 4 = 0$$ Solving for x: $$x= 2$$ So, the integrand is positive when \(x>2\). Therefore, we have two intervals to consider: \([1, 2]\) for the negative part and \([2, 6]\) for the positive part of the integrand.
02

Identify the functions for each interval

Since the integrand is the absolute value of the function \(f(x) = 2x - 4\), we already know how the function looks like in each interval. In the interval \([1, 2]\), the function is negative, so its absolute value will transform it into a positive value: \(|f(x)| = 4 - 2x\). On the other hand, in the interval \([2, 6]\), the function is already positive, so the absolute value doesn't change it: \(|f(x)| = 2x - 4\).
03

Calculate the areas under the curves for each interval

On the interval \([1, 2]\), the function \(|f(x)| = 4 - 2x\) is a line with a negative slope. Therefore, the area under this line will be the area of a triangle. Using the base and the height: $$\text{Area from 1 to 2} = \frac{1}{2} \times (2-1) \times (4 - 4)$$ On the interval \([2, 6]\), the function \(|f(x)| = 2x - 4\) is also a line, but with a positive slope. The area under this line will also be the area of a triangle. Using the base and the height: $$\text{Area from 2 to 6} = \frac{1}{2} \times (6-2) \times (8-4)$$
04

Add the areas and evaluate the integral

Now we just need to add these two areas together to get the total area under the curve of the absolute value function: $$\int_{1}^{6}|2x - 4| dx = \text{Area from 1 to 2} + \text{Area from 2 to 6}$$ Plugging in the values of the areas calculated in step 3, we have: $$\int_{1}^{6}|2x - 4| dx = \frac{1}{2} \times (2-1) \times (4 - 4) + \frac{1}{2} \times (6-2) \times (8-4) = 0 + \frac{1}{2} \times 4 \times 4 $$ So the final answer is: $$\int_{1}^{6}|2x - 4| dx = 8$$

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Key Concepts

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

Understanding the Area Under the Curve
The area under a curve in geometry represents the integral of a function. This area signifies the accumulated 'value' from one point to another. In context, here we are evaluating the integral of an absolute value function, which represents the total area bounded by the curve and the x-axis within specified limits. In simple terms, if you imagine tracing a shape below or above the x-axis, the area enclosed by those boundaries is what we are calculating. When the function is above the x-axis, it's straightforward; area can simply be counted as positive. However, when it's below, we still need to treat it as a positive contribution. This is because area can never be negative; the absolute value function assists here by flipping any negative sections above the x-axis.
Diving into the Absolute Value Function
The absolute value function, represented by \(|f(x)|\), is crucial in turning negative values into positive contributions for our area.In our exercise, the function is \(2x - 4\) under the absolute value. When \((2x - 4)\) becomes positive, the function remains unchanged.However, for values where \((2x - 4) < 0\), the absolute value flips it, maintaining positivity.Here鈥檚 how it works:
  • For \(x > 2\), \(2x - 4\) is positive, so \(|2x - 4| = 2x - 4\).
  • For \(x < 2\), \(2x - 4\) is negative, so \(|2x - 4| = -(2x - 4) = 4 - 2x\).
This flip makes the entire computation process possible because it allows us to calculate areas using geometric shapes, like triangles.
Integrating for the Full Picture
An integral is a mathematical tool that provides us with a precise way of calculating the area under curves.It brings together a sum of infinitely small areas under the curve between two points, yielding us the 'total area.'In our task, we are finding the integral of \(|2x - 4|\) from \(x = 1\) to \(x = 6\).This requires us to break down the function using interval analysis, allowing us to consider each part separately.Ultimately, we find two different triangles - for ranges \([1, 2]\) and \([2, 6]\) - compute their individual areas, and sum them to find the integral's value. This is the beauty of integration; it lets us understand the cumulative effect over a specified range, even when the function changes nature within that range.
Understanding Interval Analysis
Interval analysis is the process of breaking down a problem into manageable parts over specific ranges.This approach is beneficial when dealing with functions like absolute value expressions which have different behaviors over different intervals.For the integral \(\int_{1}^{6}|2x - 4| dx\), we dissect it around the critical point: \(x = 2\), where the inner function changes sign.This enables us:
  • Identify the behavior in each part, \([1, 2]\) where the original function \(2x - 4\) is negative.
  • Alter our function appropriately to \(|2x - 4| = 4 - 2x\) to ensure positivity, simplifying computations.
  • Analyse interval \([2, 6]\) where the function remains positive and so does not require alteration.
Using these divided ranges provides clarity, ensures each part is evaluated correctly, and when combined, give the total integral value accurately.

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Most popular questions from this chapter

Use the Substitution Rule to prove that $$\begin{array}{l} \int \sin ^{2} a x d x=\frac{x}{2}-\frac{\sin (2 a x)}{4 a}+C \text { and } \\\ \int \cos ^{2} a x d x=\frac{x}{2}+\frac{\sin (2 a x)}{4 a}+C \end{array}$$

Shape of the graph for right Riemann sums Suppose a right Riemann sum is used to approximate the area of the region bounded by the graph of a positive function and the \(x\) -axis on the interval \([a, b] .\) Fill in the following table to indicate whether the resulting approximation underestimates or overestimates the exact area in the four cases shown. Use a sketch to explain your reasoning in each case. $$\begin{array}{|l|l|l|}\hline & \text { Increasing on }[a, b] & \text { Decreasing on }[a, b] \\\\\hline \text { Concave up on }[a, b] & & \\\\\hline \text { Concave down on }[a, b] & & \\\\\hline\end{array}$$

General results Evaluate the following integrals in which the function \(f\) is unspecified. Note that \(f^{(p)}\) is the pth derivative of \(f\) and \(f^{p}\) is the pth power of \(f .\) Assume \(f\) and its derivatives are continuous for all real numbers. \(\int\left(f^{(p)}(x)\right)^{n} f^{(p+1)}(x) d x,\) where \(p\) is a positive integer, \(n \neq-1\)

Functions with absolute value Use a calculator and the method of your choice to approximate the area of the following regions. Present your calculations in a table, showing approximations using \(n=16,32,\) and 64 subintervals. Make a conjecture about the limits of the approximations. The region bounded by the graph of \(f(x)=\left|25-x^{2}\right|\) and the \(x\) -axis on the interval [0,10].

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{2} \frac{2 x}{\left(x^{2}+1\right)^{2}} d x$$
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