Problem 23 - The Calculus with Analytic Geometry

Question

In Exercises 23 through 28, find the area of the region bounded by the given curve and lines. Draw a figure showing the region and a rectangular element of area. Express the measure of the area as the limit of a Riemann sum and then as a definite integral. Evaluate the definite integral by the fundamental theorem of the calculus. \(y=4 x-x^{2} ; x\) axis; \(x=1, x=3\)

Step-by-Step Solution

Answer

The area is \(\frac{22}{3}\).

Step 1: Identify the boundaries and curve

The given curve is described by the equation \(y = 4x - x^2\) and the boundaries are the x-axis, \(x=1\) , and \(x=3\).

Step 2: Draw the region and a rectangular element

Draw the parabola described by \(y = 4x - x^2\) and plot the lines \(x=1\) , \(x=3\) , and the x-axis. The region bounded is the area under the curve between \(x=1\) and \(x=3\) . Represent a typical rectangular element of width \(dx\) and height \(y = 4x - x^2\).

Step 3: Set up the Riemann sum

The area can be expressed as the sum of the areas of rectangles. The height of each rectangle is given by \(f(x) = 4x - x^2\) and the width is \(dx\) . Thus, the approximate area \(A_n\) is given by: \[A_n = \sum_{i=1}^{n} (4x_i - x_i^2) \Delta x\]

Step 4: Express as a definite integral

To find the exact area, take the limit as \(n\) approaches infinity, which transforms the Riemann sum into a definite integral: \[A = \int_{1}^{3} (4x - x^2) \,dx\]

Step 5: Evaluate the definite integral

Use the Fundamental Theorem of Calculus to evaluate the definite integral: \[A = \int_{1}^{3} (4x - x^2) \,dx\] Find the antiderivative: \(\int (4x - x^2) dx = 2x^2 - \frac{1}{3}x^3 + C\) Now evaluate the integral from \(1\) to \(3\): \[ A = \left[ 2x^2 - \frac{1}{3}x^3 \right]_{1}^{3} \] \[= \left( 2(3)^2 - \frac{1}{3}(3)^3 \right) - \left( 2(1)^2 - \frac{1}{3}(1)^3 \right) \] \[= \left( 18 - 9 \right) - \left( 2 - \frac{1}{3} \right) \] \[= 9 - \frac{5}{3} \] \[= \frac{27}{3} - \frac{5}{3} \] \[= \frac{22}{3}\]

Step 6: Conclusion

The area of the region bounded by the curve \(y=4x-x^2\) and the lines \(x=1\) , \(x=3\) , and the x-axis is \(\frac{22}{3}\).

Key Concepts

definite integral

A definite integral is a way to calculate the area under a curve within a certain interval. For our exercise, we're interested in finding the area under the curve described by the equation \( y = 4x - x^2 \)between \( x=1 \)and \( x=3 \). Instead of adding an infinite number of tiny rectangles as we do with Riemann sums, the definite integral provides a precise value.
To set this up, we represent the area as follows:\[ A = \, \, \int_{1}^{3} (4x - x^2) \,dx \]
This symbol \( \, \int \)represents the integration process, where \( 4x - x^2 \)is the function we integrate and the limits \[ 1 \]and \[ 3 \]determine the interval on the x-axis. Calculating this integral gives us the exact area beneath the curve between these two points.

Riemann sum

To understand the area under a curve, we begin with Riemann sums. This involves dividing the area into numerous thin rectangles. Each rectangle has a width \( \Delta x \)and height given by the function value at a specific point \( f(x) = 4x - x^2 \).
The approximate area \( A_n \) can be calculated as follows:\[ A_n = \, \, \sum_{i=1}^{n} (4x_i - x_i^2) \, \Delta x \]
This summation symbol, \( \sum \), means we sum up the areas of all these rectangles.
As the number of rectangles \( n \)grows larger, each rectangle becomes thinner, and the Riemann sum becomes a better approximation of the actual area.
In the limit, as \( n \)approaches infinity, the Riemann sum transforms into a definite integral, enabling us to calculate the exact area under the curve.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, two core operations in calculus. It tells us that to find the area under a curve, we can use the antiderivative of the function:
\( F(x) = \, \, \int f(x) \, dx \)
The FTC states that if \( F'(x) = f(x) \), then:
\[ \, \, \int_{a}^{b} f(x) \,dx = F(b) - F(a) \]\
In our problem, we found the antiderivative of \( f(x) = 4x - x^2 \)to be \( F(x) = 2x^2 - \, \, \frac{1}{3} x^3 \).
Using the FTC, we evaluated it from \( x=1 \)to \( x=3 \):
\[ \, \, \bigg[ 2x^2 - \, \, \frac{1}{3} x^3 \bigg]_{1}^{3} \]
This step allowed us to find the exact area beneath the curve without endless summations.

bounded region

A bounded region refers to the area enclosed by specific boundaries. In the context of our exercise, the bounded region is framed by:

• The curve \( y = 4x - x^2 \)
• The lines \( x=1 \)and \( x=3 \)
• The x-axis (where \( y=0 \) )

To visualize this, imagine the area under the curve from \( x=1 \)to \( x=3 \) and above the x-axis. This forms a closed shape with clear boundaries.
By calculating the definite integral of the function over this interval, we find the area of this bounded region.
This process ensures we only consider the area within these specified limits, resulting in the precise area we calculated as \( \frac{22}{3} \).