Problem 21 - A Graphical Approach to Precalculus with Limits

Question

Find the exact value of each integral, using formulas from geometry. Do not use a calculator. $$\int_{1}^{4}(2 x-1) d x$$

Step-by-Step Solution

Answer

The integral evaluates to 12.

Step 1: Understand the Integral

The integral \( \int_{1}^{4}(2x-1) \, dx \) requires us to find the area under the curve \( y = 2x - 1 \) from \( x = 1 \) to \( x = 4 \). This is a linear function, which forms a region with geometrical shapes likely to be a triangle or trapezoid.

Step 2: Determine the Shape and Vertices

Plot the function \( y = 2x - 1 \). Calculate the values at the bounds: at \( x = 1 \), \( y = 2(1) - 1 = 1 \) and at \( x = 4 \), \( y = 2(4) - 1 = 7 \). The line intersects (1, 1) and (4, 7). Additionally, determine where it intersects the x-axis to form a complete shape: set \( y = 0 \), so \( 0 = 2x - 1 \), giving \( x = \frac{1}{2} \) at this point.

Step 3: Identify the Geometric Shape

The line intersects the x-axis at \( x = \frac{1}{2} \) and forms a trapezoid with the line \( y = 0 \) (the x-axis), going from \( x = 1 \) to \( x = 4 \). The vertices of this trapezoid are at (1, 1), (4, 7), (4, 0), and (1, 0).

Step 4: Calculate Area of Trapezoid

The area \( A \) of a trapezoid is given by the formula: \[ A = \frac{1}{2} \times (b_1 + b_2) \times h \], where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides and \( h \) is the height. For our trapezoid, \( b_1 = 1 \), \( b_2 = 7 \), and \( h = 4 - 1 = 3 \). Substitute these values to find the area: \[ A = \frac{1}{2} \times (1 + 7) \times 3 = \frac{1}{2} \times 8 \times 3 = 12. \]

Step 5: Conclusion

The area under the curve \( y = 2x - 1 \) from \( x = 1 \) to \( x = 4 \) is the area of the trapezoid, which equals 12. Therefore, the value of the integral is 12.

Key Concepts

geometry

Geometry is a branch of mathematics that deals with shapes, sizes, and the properties of space. In the context of integrals, geometry helps us understand the visual aspect of the problem. By looking at the graph of a function, we can determine the area beneath it using geometrical shapes such as rectangles, triangles, or trapezoids. It's like taking a 2D image and interpreting from it a specific numerical value based on its shape and dimensions. For the linear function in our original exercise, the task is to visually analyze and predict which geometrical shape best fits the space under the curve. This understanding leads us to solve the integral without doing much calculus, by simply applying geometric formulas.

area under the curve

The concept of the area under a curve is instrumental in calculus and geometry. It refers to the total space between the curve of a function and a specified segment of the x-axis. This area can be calculated using integrals, which sum up infinitesimally small rectangles under the curve. However, for certain functions, such as linear ones, we can use geometric methods to find this area quickly. In our specific example, the curve is defined by the function \(y = 2x - 1\). The integral from \(x = 1\) to \(x = 4\) gives us the exact area under this line segment. By knowing the shape of the region (a trapezoid) formed by this function, we can directly calculate the area using a simple trapezoid formula instead of standard integration techniques.

trapezoid area

The area of a trapezoid is a fundamental geometric concept that can be applied to solve integrals involving linear functions. A trapezoid, also known as a trapezium in some countries, is a quadrilateral with at least one pair of parallel sides. To find the area of a trapezoid, we use the formula:

• Identify the bases \(b_1\) and \(b_2\), which are the two parallel sides.
• Determine the height \(h\), perpendicular distance between the bases.

For our trapezoid, formed by the line \(y = 2x - 1\) and the x-axis between \(x = 1\) and \(x = 4\), the bases are \(b_1 = 1\) and \(b_2 = 7\), calculated from the y-values of the line at the bounds. The height \(h\) is the horizontal distance, \(3\), between these bounds. An area calculation using the formula \(A = \frac{1}{2} (b_1 + b_2) \times h\) yields \(12\), which is both the area under the curve and the value of the integral. Understanding this process within the realm of geometry simplifies the concept of integration.