Problem 10 - Calculus: Early Transcendentals

Question

Use the Midpoint Rule with the given value of \( n \) to approximate the integral. Round the answer to four decimal places. \( \displaystyle \int^1_0 \sqrt{x^3 + 1}\, dx \), \( n = 5 \)

Step-by-Step Solution

Answer

The approximate integral using the Midpoint Rule is 1.0671.

Step 1: Understand the Midpoint Rule

The Midpoint Rule is a method for approximating the integral of a function using the average value of the function at the midpoint of each subinterval. The formula for the Midpoint Rule is \( M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x \), where \( \Delta x = \frac{b-a}{n} \) and \( \bar{x}_i \) is the midpoint of the \( i \)-th subinterval.

Step 2: Calculate the Width of Each Subinterval

Given the interval \([0, 1]\) and \( n = 5 \), calculate \( \Delta x \):\[ \Delta x = \frac{b-a}{n} = \frac{1 - 0}{5} = 0.2 \].

Step 3: Determine the Midpoints

Determine the midpoints \( \bar{x}_i \) for each subinterval:- \( \bar{x}_1 = 0.1 \)- \( \bar{x}_2 = 0.3 \)- \( \bar{x}_3 = 0.5 \)- \( \bar{x}_4 = 0.7 \)- \( \bar{x}_5 = 0.9 \)

Step 4: Evaluate the Function at Each Midpoint

Evaluate \( f(x) = \sqrt{x^3 + 1} \) at each midpoint:- \( f(\bar{x}_1) = \sqrt{0.1^3 + 1} = \sqrt{1.001} \)- \( f(\bar{x}_2) = \sqrt{0.3^3 + 1} = \sqrt{1.027} \)- \( f(\bar{x}_3) = \sqrt{0.5^3 + 1} = \sqrt{1.125} \)- \( f(\bar{x}_4) = \sqrt{0.7^3 + 1} = \sqrt{1.343} \)- \( f(\bar{x}_5) = \sqrt{0.9^3 + 1} = \sqrt{1.729} \)

Step 5: Approximate the Integral using the Midpoint Rule

Using the Midpoint Rule formula, calculate the approximation: \( M_5 \approx 0.2 \left(\sqrt{1.001} + \sqrt{1.027} + \sqrt{1.125} + \sqrt{1.343} + \sqrt{1.729}\right) \).Calculate the expression to find the sum:- Sum of \( f(\bar{x}_i) \) values: \( \approx \sqrt{1.001} + \sqrt{1.027} + \sqrt{1.125} + \sqrt{1.343} + \sqrt{1.729} \approx 5.3354 \).- Multiply by \( \Delta x = 0.2 \) to get: \( M_5 \approx 0.2 \times 5.3354 \approx 1.0671 \).

Key Concepts

Numerical Integration

In calculus, numerical integration is a method for estimating the value of definite integrals. Integrals can be challenging to calculate analytically, especially when the function is complex. Numerical methods simplify this by using algorithms to find approximate solutions. One popular technique is the Midpoint Rule, which falls under this broader category.

• Integrals provide the area under a curve or mathematical function, and numerical integration helps find this area quantitatively.
• Numerical integration is essential for solving real-world problems where exact solutions are difficult to find.
• Methods like the Midpoint Rule, Trapezoidal Rule, and Simpson's Rule are all numerical integration techniques.

By employing these techniques, students and professionals can tackle complex integrals without requiring advanced analytical methods.

Approximate Integral

Approximating an integral involves estimating the area under a curve when an exact solution is difficult to find. When using the Midpoint Rule, we achieve this by calculating the height of the function at the center of each subinterval.

• The Midpoint Rule uses the average value of the function at these center points to estimate the entire area.
• This method is particularly effective for functions that are reasonably smooth over the interval.
• Approximating integrals allows for practical calculations in physics, engineering, and statistics.

While the Midpoint Rule provides an estimate, it becomes more accurate as the number of subintervals increases.

Subinterval Midpoints

Subinterval midpoints are crucial in using the Midpoint Rule for numerical integration. These are the points within smaller segments of the overall interval, ensuring we capture the function's behavior precisely.
To calculate the midpoints for an interval \( [a, b] \), divide it into \( n \) equal parts, and then find the center of each part:

• Formula for midpoint: \[ \bar{x}_i = a + \left(i - \frac{1}{2}\right) \Delta x \]
• This calculation provides the points where the function should be evaluated.
• Using these midpoints helps accurately approximate the overall integral by assessing the function's value at these strategic points.

Understanding these midpoints is key to getting better results from the numerical integration process.

Calculus Techniques

Calculus techniques, such as the Midpoint Rule, offer practical tools for solving integral calculus problems. These methods, like the Midpoint Rule, break complex integrals into manageable parts.

• Integrals represent accumulated quantities, such as area, volume, and other measures.
• Calculus techniques apply rules and formulas to solve these integrals effectively.
• By understanding and utilizing these techniques, students can handle various calculus problems with confidence.

In particular, the Midpoint Rule, as a calculus technique, highlights how mathematical concepts can be applied to find useful approximate solutions.