Question

Draw the graph of \(\mathrm{f}\), and evaluate \(\int_{a}^{b} \mathrm{f}(\mathrm{x}) \mathrm{dx}\) in each of the following examples. (a) \(\mathrm{f}(\mathrm{x})= \begin{cases}1 & \text { if }-\infty<\mathrm{x} \leq 0 \\ 5 & \text { if } 0<\mathrm{x}<2, \text { and }[\mathrm{a}, \mathrm{b}]=[-3,3] \\ 3 & \text { if } 2 \leq \mathrm{x}<\infty\end{cases}\) (b) \(f(x)=\left\\{\begin{array}{cc}x^{2} & \text { if }-\infty

Short answer

Question: Calculate the definitive integrals for the following functions over the given intervals: (a) A piecewise function defined by three separate cases: (1) \(f(x) = 1\) for \(-\infty < x \leq 0\), (2) \(f(x) = 5\) for \(0 < x < 2\), and (3) \(f(x) = 3\) for \(2 \leq x < \infty\). The interval of integration is [-3, 3]. (b) A function defined by two cases: \(f(x) = x^2\) for \(-\infty < x < 0\) and \(f(x) = 2 - x^2\) for \(0 \leq x < \infty\). The interval of integration is [-2, 2]. (c) A function that takes the value of the integer n for all x greater than or equal to n but less than n+1. The interval of integration is [0, 5]. Answer: (a) The definitive integral for the piecewise function a over the interval [-3, 3] is 16. (b) The definitive integral for the function b over the interval [-2, 2] is -\(\frac{4}{3}\). (c) The definitive integral for the function c over the interval [0, 5] is 10.

Step-by-step solution

Graph the function f(x) in the given range

To graph the function, we will look at the three different cases defined for this piecewise function: - Case 1: When \(-\infty<x\leq 0\), \(f(x) = 1\). In this interval, the graph will be a horizontal line at a height of 1. - Case 2: When \(0<x<2\), \(f(x) = 5\). In this interval, the graph will also be a horizontal line but at a height of 5. - Case 3: When \(2 \leq x < \infty\), \(f(x) = 3\). In this interval, the graph will be a horizontal line at a height of 3.

Evaluate the integral

To evaluate the integral \(\int_{a}^{b} f(x) \, dx\), we will divide it into three separate integrals according to the subintervals defined in the piecewise function. The given interval is [-3, 3]: - Integral 1: \(\int_{-3}^{0} 1 \, dx\): This integral can be found by finding the area of the rectangle formed with length 3 (from -3 to 0 in the x-axis) and height 1. So, the value of this integral is \((3-(-3)) \cdot 1 = 3\). - Integral 2: \(\int_{0}^{2} 5 \, dx\): This integral is the area of the rectangle formed with length 2 (from 0 to 2 in the x-axis) and height 5. So, the value of this integral is \((2-0) \cdot 5 = 10\). - Integral 3: \(\int_{2}^{3} 3 \, dx\): This integral is the area of the rectangle formed with length 1 (from 2 to 3 in the x-axis) and height 3. So, the value of this integral is \((3-2) \cdot 3 = 3\). The definite integral \(\int_{a}^{b} f(x) dx\) from -3 to 3 is equal to the sum of the three individual integrals previously computed: \(\int_{-3}^{3} f(x) \, dx = 3 + 10 + 3 = 16\) (b)

Graph the function f(x) in the given range

To graph the function, we will look at the two different cases defined for this function: - Case 1: When \(-\infty<x<0\), \(f(x) = x^2\). In this interval, the graph is a parabola opening upwards. - Case 2: When \(0\leq x<\infty\), \(f(x) = 2 - x^2\). In this interval, the graph is a parabola opening downwards with its vertex at x=0.

Evaluate the integral

The given interval for this function is [-2, 2], and we divide the integral into two parts: - Integral 1: \(\int_{-2}^{0} x^2 \, dx\): For this integral, we use the power rule, which says that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where n is a constant and C is the constant of integration. So, here, `\(\int_{-2}^{0} x^2 dx = \frac{x^{2+1}}{2 + 1} \Big |_{-2}^{0} =\frac{x^3}{3} \Big |_{-2}^{0} = \frac{(0)^3}{3} - \frac{(-2)^3}{3} = 0 - \frac{8}{3} = -\frac{8}{3}\). - Integral 2: \(\int_{0}^{2} (2-x^2) \, dx\): For this integral, we break it into two terms and integrate separately: 1. \(\int_{0}^{2} 2 \, dx\): Using the power rule, we get \(\int_{0}^{2} 2 \, dx = 2x \Big |_{0}^{2} = 2(2) - 2(0) = 4\). 2. \(\int_{0}^{2} -x^2 dx\): Using the power rule, we get `\(\int_{0}^{2} -x^2 dx = -\frac{x^3}{3} \Big |_{0}^{2} = -\frac{(2)^3}{3} - (-\frac{(0)^3}{3}) = -\frac{8}{3}\)`. Combining the two integrals, we get: \(\int_{-2}^{2} f(x) \, dx = -\frac{8}{3} + 4 -\frac{8}{3} = -\frac{16}{3} + 4 = -\frac{4}{3}\). (c)

Graph the function

In this step, we graph the function f(x) = n for all x belonging to the interval between n and n+1, for integer values of n. The graph consists of horizontal line segments at heights equal to integer values of n, with each segment connecting (n, n) and (n+1, n).

Evaluate the integral

We are given the interval [0, 5]. The function f(x) takes the value of an integer where x is greater than or equal to that integer but less than the integer plus one. So, we need to divide the integration into separate integrals for each integer value of n from 0 to 4: - Integral 1: \(\int_{0}^{1} 0 \, dx\): The value of this integral is 0 because the function is 0 for x between 0 and 1. - Integral 2: \(\int_{1}^{2} 1 \, dx\): The value of this integral is 1 because the function is 1 for x between 1 and 2. - Integral 3: \(\int_{2}^{3} 2 \, dx\): The value of this integral is 2 because the function is 2 for x between 2 and 3. - Integral 4: \(\int_{3}^{4} 3 \, dx\): The value of this integral is 3 because the function is 3 for the x between 3 and 4. - Integral 5: \(\int_{4}^{5} 4 \, dx\): The value of this integral is 4 because the function is 4 for x between 4 and 5. Adding the values of the above integrals, we get: \(\int_{0}^{5} f(x) \, dx = 0 + 1 + 2 + 3 + 4 = 10\)

Key concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics concerned with the process of integration, which is the inverse operation of differentiation. It deals with the accumulation of quantities, such as areas under curves and the net change of a function over an interval. A definite integral, symbolized by \(\int_{a}^{b} f(x) \, dx\), calculates the accumulated value of a function \(f(x)\) from the lower limit \(a\) to the upper limit \(b\). The evaluation of a definite integral results in a numerical value which represents the total accumulation of the function between two points. This concept is extensively used to solve real-world problems involving displacement, area, volume, and total growth, among others.

The evaluation of the definite integral often requires identifying the type of function – whether it's a continuous function, piecewise function, or otherwise – and using the appropriate integration techniques to find the integral's value.

Piecewise Functions

Piecewise functions are defined by different expressions over different parts of their domain. They often represent real-world situations where a certain rule or relationship changes based on the input. For example, a cost function that varies depending on the quantity purchased can be represented by a piecewise function.

Graphing piecewise functions involves plotting these different expressions over their respective intervals, ensuring that the graph is a true reflection of the function's definition over its entire domain. When it comes to integrating a piecewise function, the process requires evaluating a separate integral for each section of the function's domain and then summing these individual results to find the total integral over the desired interval. This segmented approach allows for the accurate calculation of areas, even when the function behaves differently across subintervals.

Integrals of Continuous Functions

A continuous function is one that is unbroken and has no gaps, jumps, or holes throughout its domain. The definite integral of a continuous function over an interval can be interpreted as the net area between the function's graph and the x-axis, within the interval limits.

When calculating the definite integral of a continuous function, you can generally apply standard integration techniques, such as the power rule, integration by parts, or substitution. For continuous piecewise functions, while the function overall might have jump discontinuities between pieces, each segment of the function to be integrated must be continuous within its subinterval. Care must be taken to ensure that these boundaries of integration align with the points where the function's expression changes.

Graphing Functions

Graphing functions involves visually representing a function's behavior on a coordinate system. For functions that are not piecewise, this usually means plotting a smooth curve or line that matches the function's formula. For piecewise functions and definite integrals, the graph can assist in visualizing the area being calculated under the curve.

The accurate graph representation of functions is crucial in integral calculus as it helps to conceptualize the areas to be integrated, especially with piecewise functions that change behavior at defined intervals. By sketching the function within the relevant interval, students can better understand how to partition the interval for integration and how to apply different integration strategies to various parts of the function.