Problem 56 - Calculus

Question

Find the area of the region under the curve \(y=f(x)\) over the interval \([a, b] .\) To do this, divide the interval \([a, b]\) into \(n\) equal subintervals, calculate the area of the corresponding circumscribed polygon, and then let \(n \rightarrow \infty\). (See the example for \(y=x^{2}\) in the text.) \(y=x^{2} ; a=-2, b=2\)

Step-by-Step Solution

Answer

The area is \(\frac{16}{3}\).

Step 1: Define the Function and Interval

We are given a function \( y = x^2 \) and we need to find the area under this curve from \( a = -2 \) to \( b = 2 \). This means we're looking for the area under the parabola over the interval [-2, 2].

Step 2: Divide the Interval into Subintervals

We divide the interval \([-2, 2]\) into \(n\) subintervals of equal width. The width of each subinterval, \( \Delta x \), is given by \( \Delta x = \frac{b - a}{n} = \frac{2 - (-2)}{n} = \frac{4}{n} \).

Step 3: Determine the Height of Each Rectangle

The height of each rectangle is determined by the value of the function at a specific point within the subinterval. We'll use the right endpoint of each subinterval, \( x_i = -2 + i \Delta x \), where \( i \) ranges from 1 to \( n \). Therefore, the height of the rectangles is \( f(x_i) = x_i^2 = (-2 + i\Delta x)^2 \).

Step 4: Calculate the Area of Each Rectangle

The area of each rectangle is the product of its height and width. Therefore, the area is \( f(x_i) \times \Delta x = (-2 + i\Delta x)^2 \times \frac{4}{n} \).

Step 5: Sum the Areas of All Rectangles

The total area under the curve is approximated by summing the areas of all rectangles: \[A_n = \sum_{i=1}^{n} \left((-2 + i\Delta x)^2 \cdot \frac{4}{n}\right) \]. Substituting \( \Delta x = \frac{4}{n} \), the sum becomes: \[A_n = \sum_{i=1}^{n} \frac{4}{n}(-2 + \frac{4i}{n})^2 \].

Step 6: Take the Limit as n Approaches Infinity

To find the exact area under the curve, we take the limit as \( n \rightarrow \infty \): \[A = \lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}\left((-2 + \frac{4i}{n})^2\right) \]. This limit can be calculated using the definite integral: \[\int_{-2}^{2} x^2 \, dx \].

Step 7: Integrate to Find the Exact Area

We compute the definite integral: \[\int_{-2}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \left(\frac{2^3}{3} - \frac{(-2)^3}{3}\right) = \left(\frac{8}{3} - \frac{-8}{3}\right) = \frac{16}{3} \].

Step 8: Conclusion

The area under the curve \( y = x^2 \) from \( x = -2 \) to \( x = 2 \) is \( \frac{16}{3} \).

Key Concepts

Riemann Sums

Riemann sums are a fundamental tool in calculus that help us approximate the area under a curve. Imagine dividing the interval from \(a\) to \(b\) into \(n\) smaller sections. For each section, we create a rectangle that touches the curve at one endpoint, usually choosing the right endpoint for simplicity. The height of each rectangle is determined by the function's value at that point, making the area of each rectangle the product of this height and the width of the subinterval.

The total area is then approximated by adding up these rectangular areas. Mathematically, we express this sum as \( A_n = \sum_{i=1}^{n} f(x_i) \Delta x \). As \(n\) becomes very large, the width of each rectangle becomes very small, leading to a more accurate approximation of the area under the curve. This process leads us towards the concept of an integral.

• Divide interval \([a, b]\) into \(n\) subintervals.
• For each subinterval, select a point, typically the right endpoint.
• Use this point to determine the height of the rectangle.
• Calculate the area by multiplying height and width.
• Sum all rectangular areas to approximate total area.

Area Under a Curve

The area under a curve for a given function, on a specified interval, represents the definite integral of that function. The concept of the area under a curve is essential in calculus because it provides a way to compute quantities that accumulate continuously. For instance, it allows us to find the total area between a curve and the x-axis over an interval \([a, b]\).

In our exercise, we use the interval \([-2, 2]\) for the curve \(y = x^2\). By computing the definite integral of the function, we obtain the exact area. The process involves setting up a Riemann sum, similar to adding up the areas of a set of rectangles, and then taking the limit as the number of rectangles approaches infinity. This limit turns into the integral \( \int_{-2}^{2} x^2 \; dx \).

• Integral of \(f(x)\) from \(a\) to \(b\) gives exact area below the curve.
• Converts approximation (Riemann sums) into precise mathematical calculation.
• Critical for calculating accumulated values.

Polynomial Functions

Polynomial functions are expressions involving terms with non-negative integer exponents on the variable \(x\). They look like \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\). These functions are smooth and continuous, making them easier to handle in calculus.

In the exercise, \(y = x^2\) is a simple polynomial function with one non-zero term, \(x^2\). To find the area under such curves, we use calculus tools like integration because they provide exact answers. Polynomial functions, especially quadratics like \(y = x^2\), are fundamental in learning about integrals and calculus as a whole. They serve as a great introduction due to their straightforward patterns and predictable behavior.

• Polynomial functions consist of sums of power functions.
• They are continuous and differentiable, ideal for calculus operations.
• Quadratic functions \((ax^2 + bx + c)\) are common and easy to integrate.