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Chapter 4: Problem 21

You are given a definite integral \(\int_{a}^{b} f(x) d x\). Make a sketch of fon \([a, b] .\) Then use the geometric interpretation of the integral to evaluate it. $$ \int_{-2}^{4} 3 d x $$

Short Answer

Expert verified
The definite integral \(\int_{-2}^{4} 3 dx\) represents the area under the curve of the function \(f(x) = 3\) on the interval \([-2,4]\). This forms a rectangle with width \(6\) and height \(3\). The area of the rectangle is \(A = 6 \times 3 = 18\), so the definite integral evaluates to \(18\).

Step by step solution

01

Sketch the function

The given function is \(f(x) = 3\), a constant function. To sketch it, we'll draw a horizontal line at \(y = 3\) on our coordinate plane. The interval to consider is \([-2,4]\) on the x-axis.
02

Determine the shape formed by the function and the interval

On the interval \([-2,4]\), the function creates a rectangle with the base on the x-axis and the top along the line \(y=3\). The vertices of the rectangle are \((-2,0)\), \((-2,3)\), \((4,3)\), and \((4,0)\).
03

Determine the dimensions of the rectangle

Since the function is a line at \(y=3\) and the interval spans from \(-2\) to \(4\) on the x-axis, we can easily determine the dimensions of the rectangle that is formed. The width will be from \(-2\) to \(4\), which is a total width of \((4 - (-2)) = 6\) units. The height is given by the function value, which is always \(3\).
04

Calculate the area of the rectangle

Now, knowing the dimensions of the rectangle, we can calculate its area. The area of a rectangle is given by the formula \(A = width \times height\). So, in this case, we have: $$ A = 6 \times 3 $$
05

Evaluate the definite integral

Since the area under the curve of the function \(f(x) = 3\) on the interval \([-2,4]\) is represented by the definite integral, we can write that: $$ \int_{-2}^{4} 3 dx = A = 6 \times 3 = 18 $$ So the definite integral \(\int_{-2}^{4} 3 dx\) evaluates to \(18\).

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Key Concepts

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

Geometric Interpretation of Integrals
The geometric interpretation of integrals is an approach that allows us to understand definite integrals in terms of area. When we consider the integral \(\int_{a}^{b} f(x) dx\), it can be visualized as the net area enclosed between the graph of \(f(x)\) and the x-axis, from \(x=a\) to \(x=b\).

If the function \(f(x)\) lies above the x-axis in this interval, the integral gives the area of the region. If \(f(x)\) lies below the x-axis, the integral accounts for the area with a negative sign, which can be interpreted as a 'deficit', compared to being above the x-axis. In our exercise, we consider a constant function \(f(x) = 3\) over the interval \(x = -2\) to \(x = 4\), which creates a rectangle above the x-axis. The definite integral, in this case, will yield the area of this rectangle, illustrating how integrals can represent geometric area in a straightforward manner.
Area Under Curve
The concept of 'area under curve' provides a visual and intuitive grasp of definite integrals. When we compute \(\int_{a}^{b} f(x) dx\), we are essentially adding up infinitely small 'slices' of area under the curve of \(f(x)\) between the points \(a\) and \(b\).

In our example, because \(f(x)\) is a constant function, the 'curve' is a straight line, and the area is a simple rectangular shape, making the integral's evaluation particularly simplified. The base of this rectangle corresponds to the interval on the x-axis (from \(x=-2\) to \(x=4\)), and the height is the constant value of the function (\(y=3\)). The area represents a clear and concrete quantity that we can calculate using the formula for the area of a rectangle, and this demonstrable concept anchors the more abstract notion of integration to something tangible.
Constant Function
A constant function is a type of function where the value of \(f(x)\) is the same for every input \(x\). Graphically, it's represented by a horizontal line on a coordinate plane. In our exercise, \(f(x) = 3\), which means no matter what \(x\)-value we input, the output will always be \(3\).

This simplicity actually provides a perfect instance for understanding concepts in calculus because the area under a constant function's curve over any interval is easily calculated as a rectangle. The constant nature of the function means that when we evaluate the definite integral, we're ensuring the result is the product of the interval's length and the function's constant value. Such clear-cut cases are great for solidifying a student's comprehension of the fundamental principles of calculus.

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Most popular questions from this chapter

The production of oil (in millions of barrels per day) extracted from oil sands in Canada is projected to be $$P(t)=\frac{4.76}{1+4.11 e^{-0.22 t}} \quad 0 \leq t \leq 15$$ where \(t\) is measured in years, with \(t=0\) corresponding to the beginning of \(2005 .\) What will the total oil production of oil from oil sands be over the years from the beginning of 2005 until the beginning of \(2020(t=15)\) ?

Let \(f\) be continuous on \((-\infty, \infty)\), and let \(c\) be a constant. Show that $$ \int_{c a}^{c b} f(x) d x=c \int_{a}^{b} f(c x) d x $$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. Assuming that the integral exists and that \(f\) is even and \(g\) is odd, then $$ \int_{-a}^{a} f(x)[g(x)]^{2} d x=2 \int_{0}^{a} f(x)[g(x)]^{2} d x $$

The number of hours of daylight at any time \(t\) in Chicago is approximated by $$L(t)=2.8 \sin \left[\frac{2 \pi}{365}(t-79)\right]+12$$ where \(t\) is measured in days and \(t=0\) corresponds to January 1 . What is the daily average number of hours of daylight in Chicago over the year? Over the summer months from June \(21(t=171)\) through September \(20(t=262)\) ?

According to data from the American Petroleum Institute, the U.S. Strategic Petroleum Reserves from the beginning of 1981 to the beginning of 1990 can be approximated by the function $$ S(t)=\frac{613.7 t^{2}+1449.1}{t^{2}+6.3} \quad 0 \leq t \leq 9$$ where \(S(t)\) is measured in millions of barrels and \(t\) in years, with \(t=0\) corresponding to the beginning of 1981 . Using the Trapezoidal Rule with \(n=9\), estimate the average petroleum reserves from the beginning of 1981 to the beginning of 1990 .
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