Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

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 \(n \rightarrow \infty\). $$ y=2 x+2 ; a=-1, b=1 . \text { Hint: } x_{i}=-1+\frac{2 u}{n} $$

Step-by-Step Solution

Answer

The area is 6 square units.

Step 1: Understand the Problem

We need to find the area under the curve described by the function \( y = 2x + 2 \) from \( a = -1 \) to \( b = 1 \). We achieve this by using the concept of integration, which involves dividing the interval \([-1, 1]\) into \(n\) equal parts and calculating the limit of the sum of areas as \(n\rightarrow \infty\).

Step 2: Divide the Interval into Subintervals

The interval \([-1, 1]\) is divided into \(n\) equal parts using the formula \( x_i = -1 + \frac{2i}{n} \), where \(i\) ranges from 0 to \(n\). This formula determines the endpoints of each subinterval.

Step 3: Express the Subinterval Width

The width of each subinterval, denoted as \( \Delta x \), can be calculated as \( \Delta x = \frac{2}{n} \) since the entire interval length from \(-1\) to \(1\) is 2.

Step 4: Determine the Height of Rectangles

The height of each rectangle can be calculated using the function's value at the right endpoint of each subinterval: \( f(x_i) = 2x_i + 2 \). For the \(i^{th}\) subinterval, the height is \( f\left(-1 + \frac{2i}{n}\right) = 2\left(-1 + \frac{2i}{n}\right) + 2 \).

Step 5: Calculate the Riemann Sum

The approximate area under the curve can be calculated using the Riemann sum: \[ S_n = \sum_{i=1}^{n} f\left(-1 + \frac{2i}{n}\right) \cdot \Delta x = \sum_{i=1}^{n} \left(2\left(-1 + \frac{2i}{n}\right) + 2\right) \cdot \frac{2}{n} \]. Simplify this expression to find the sum of areas of the rectangles.

Step 6: Compute the Limit as n Approaches Infinity

To find the exact area, calculate the limit of \( S_n \) as \( n \rightarrow \infty \). This is equivalent to evaluating the definite integral of \( 2x + 2 \) from \(-1\) to \(1\), which is: \[ \int_{-1}^{1} (2x + 2) \, dx \]. The integration yields: \[ \left. \left( x^2 + 2x \right) \right|_{-1}^{1} = \left(1^2 + 2\cdot1\right) - \left((-1)^2 + 2\cdot(-1)\right) = (1 + 2) - (1 - 2) = 6 \].

Step 7: Conclude the Calculation

The area under the curve \( y = 2x + 2 \) from \( x = -1 \) to \( x = 1 \) is 6 square units.

Key Concepts

Definite Integral

In mathematics, the definite integral is a fundamental concept used to calculate the area under a curve between two specific points on the x-axis, known as the limits of integration. For the function \( y = 2x + 2 \) over the interval \([-1, 1]\), the definite integral is evaluated as \( \int_{a}^{b} f(x) \, dx \), where \( a = -1 \) and \( b = 1 \).
The process of finding a definite integral involves calculating the antiderivative (also known as the indefinite integral) of the function and then applying the Fundamental Theorem of Calculus.

• The antiderivative of \( 2x + 2 \) is \( x^2 + 2x \).
• Applying the limits, the calculation becomes \( \left.\left(x^2 + 2x\right)\right|_{-1}^{1} \).
• The solution shows that \( (1^2 + 2 \cdot 1) - ((-1)^2 + 2 \cdot (-1)) = 6 \), thus, the area under the curve is 6 square units.

Using definite integrals provides an exact measure of area, unlike approximations obtained through other methods like Riemann sums.

Riemann Sum

Riemann sums are a way to approximate the area under a curve using a finite sum of areas of rectangles. This method helps visualize the concept of integration by converting curves which are difficult to integrate directly into manageable geometries.
To find the Riemann sum, the interval \([-1, 1]\) is split into \(n\) equal parts, and the area is given by the sum \( S_n = \sum_{i=1}^{n} f(x_i) \Delta x \).

• The width of each rectangle, \( \Delta x \), is \( \frac{2}{n} \).
• The height of each rectangle is determined by evaluating the function \( y = 2x + 2 \) at the right endpoint of each subinterval.
• This formula: \( 2(-1 + \frac{2i}{n}) + 2 \), generates these heights.

As \( n \to \infty \), the approximation becomes perfectly accurate, leading to the exact area as computed by the definite integral.

Area Under a Curve

The concept of the area under a curve is central to understanding integration. It signifies the region bounded by the curve, the x-axis, and the vertical lines at \( x = a \) and \( x = b \). This area is calculated using definite integrals.
For our function \( y = 2x + 2 \), the area under the curve from \( x = -1 \) to \( x = 1 \) can be visualized graphically. The line forms a sloping rectangle-like region above the x-axis.

• The endpoints \( x = -1 \) and \( x = 1 \) are the boundaries for this calculation.
• The linear nature of \( y = 2x + 2 \) results in a geometric figure that suggests a straightforward computation of area.

Intuitively, the area under the curve provides a real-world context by representing quantities like distance traveled or total accumulation over time, depending on the function and its units.

Subintervals

Subintervals are smaller sections into which a main interval is divided when using methods like Riemann sums to calculate areas. For the definite integral over the interval \([-1, 1]\), splitting this into \(n\) subintervals allows for smaller, more precise approximations of the area under the curve.
Each subinterval \( [x_{i-1}, x_i] \) is defined with endpoints calculated as \(x_i = -1 + \frac{2i}{n}\).

• The formula ensures equal spacing of each subinterval, making calculations uniform.
• As the number of subintervals \(n\) increases, the approximation through Riemann sums becomes more accurate.

By taking the limit as \( n \rightarrow \infty \), each of these subintervals shrinks, leading the Riemann sum to converge on the true area calculated by integration.