Problem 29 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graph and find the area of the region bounded by the graph of the function \(f\) and the lines \(y=0, x=a\), and \(x=b\) $$f(x)=-x^{2}+4 x-3 ; a=-1, b=2$$

Step-by-Step Solution

Answer

The graph of the function \(f(x)=-x^2+4x-3\) is a parabola opening downwards with x-intercepts at \((-1,0)\) and \((2,0)\), and a maximum point at \((1,0)\). The area of the region enclosed by the graph, the x-axis (\(y=0\)), and the lines at \(x=-1\) and \(x=2\) can be found using integration: $$\text{Area} = \int_{-1}^{2}f(x)\,dx=\int_{-1}^{2}(-x^2+4x-3)\,dx = 3$$ The area of the region is 3 square units.

Step 1: Sketch the graph of the function f(x)

Observe that f(x) is a quadratic function with a negative leading coefficient, so its graph will be a parabola opening downwards. To draw the graph, we also need to find the x-intercepts, where f(x) = 0: $$0=-x^{2}+4x-3$$ Use the quadratic formula to find the roots: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-4\pm\sqrt{4^2-4(-1)(-3)}}{-2}=\frac{-4\pm\sqrt{4}}{-2}$$ Thus, x-intercepts are \(x=1\pm\sqrt{1}=-1,2\). The maximum point is at \(x=-\frac{-4}{2(-1)}=1\), and the value of the function at the maximum point is \(f(1)=-1^2+4(1)-3=0\). Now, we can sketch the graph of \(f(x)\) with the x-intercepts \((-1,0)\) and \((2,0)\), and the maximum point at \((1,0)\).

Step 2: Find the area of the region

The region we are concerned with is enclosed by the graph of the function \(f(x)\), the x-axis (\(y=0\)), and the lines at \(x=-1\) and \(x=2\). We need to find the area of this region using integration. Since f(x) is a continuous function, we can use the integral to compute the area: $$\text{Area} = \int_{a}^{b}f(x)\,dx=\int_{-1}^{2}(-x^2+4x-3)\,dx$$ Now, we will evaluate the integral: $$\int (-x^2+4x-3)\,dx =\int -x^2\,dx+\int 4x\,dx-\int 3\,dx = -\frac{1}{3}x^3+2x^2-3x+C$$ Apply the limits of integration: $$\text{Area} = -\frac{1}{3}(2)^3+2(2)^2-3(2)- [-\frac{1}{3}(-1)^3+2(-1)^2-3(-1)] = 3$$ The area of the region bounded by the graph of the function and the lines is 3 square units.

Key Concepts

Area Under Curve

When we talk about the area under a curve in integral calculus, we're often interested in the space between a function and the x-axis over a particular interval. It's a visual representation of the cumulative effect or total value that the function represents over that interval.

To calculate this area, especially when it's bounded by the function and horizontal lines (typically the x-axis), we use definite integrals. The calculation involves finding the integral of the function from one boundary to the other, which in practical terms often represents the 'total accumulated value' between those points. In the case of our exercise, we are looking for the area bounded by the quadratic function, the x-axis, and the vertical lines at x=a and x=b.

Quadratic Functions

Quadratic functions are second-degree polynomial functions of the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. These functions graph to a shape known as a parabola. Depending on the sign of the leading coefficient \( a \), the parabola opens upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)).

To graph a quadratic function effectively, determining the roots (or x-intercepts) is often the first step. These are the points at which the function crosses the x-axis, found by solving \( f(x) = 0 \). Additionally, understanding the vertex of the parabola, which is the highest or lowest point of the graph, can provide a clearer picture of the function’s shape. The exercise given shows a parabola opening downwards as indicated by its negative leading coefficient.

Definite Integral

The definite integral of a function gives us the total accumulation of the function's values between two points. Graphically, it represents the net area under the curve of the function between two specific x-values, designated as the lower and upper limits of integration. For functions that lie above the x-axis in this interval, the area is positive, while functions that dip below the x-axis contribute negative area.

In mathematical notation, a definite integral is expressed as \( \[\int_a^b f(x)\,dx\] \), where \( a \) and \( b \) are the lower and upper bounds, respectively, and \( f(x) \) is the function being integrated. In our exercise, the definite integral operation quantifies the precise area between the downward-opening parabola, given by the quadratic function, and the x-axis from \( x=-1 \) to \( x=2 \).

Graphing Parabolas

A parabola is the graphical representation of a quadratic function and is characterized by its distinct 'U' shape.

To accurately graph a parabola, we need to identify several key features: the vertex, the axis of symmetry, the direction it opens (up or down), and its intercepts. The vertex can be found by using the vertex formula \( x = -\frac{b}{2a} \), which also gives us the axis of symmetry of the parabola. The intercepts include both the x-intercepts, where the graph crosses the x-axis, and the y-intercept, where it crosses the y-axis. In the exercise we are considering, once we know the parabola opens downwards (since 'a' is negative) and identify the x-intercepts and maximum point, we can sketch the graph to visualize the region whose area we are calculating.