Question

Approximate the integral to three decimal places using the indicated rule. \(\int_{0}^{3} \frac{1}{1+x^{3}} d x ;\) trapezoidal rule; \(n=6\)

Short answer

The approximate integral value using the trapezoidal rule is 1.180.

Step-by-step solution

Calculate the Width of Each Subinterval

To apply the trapezoidal rule, first determine the width of each subinterval. The formula for the width of each subinterval is \( \Delta x = \frac{b-a}{n} \), where \( a = 0 \), \( b = 3 \), and \( n = 6 \). Thus, \( \Delta x = \frac{3-0}{6} = 0.5 \).

Determine the Subinterval Endpoints

Calculate the endpoints of each subinterval using the width \( \Delta x = 0.5 \). The endpoints are \( x_0 = 0 \), \( x_1 = 0.5 \), \( x_2 = 1.0 \), \( x_3 = 1.5 \), \( x_4 = 2.0 \), \( x_5 = 2.5 \), and \( x_6 = 3.0 \).

Apply the Function to Each Endpoint

Calculate the function \( f(x) = \frac{1}{1+x^3} \) at each endpoint determined in Step 2: - \( f(x_0) = \frac{1}{1+0^3} = 1 \)- \( f(x_1) = \frac{1}{1+(0.5)^3} \approx 0.941 \)- \( f(x_2) = \frac{1}{1+1^3} = 0.5 \)- \( f(x_3) = \frac{1}{1+(1.5)^3} \approx 0.228 \)- \( f(x_4) = \frac{1}{1+2^3} = 0.111 \)- \( f(x_5) = \frac{1}{1+(2.5)^3} \approx 0.060 \)- \( f(x_6) = \frac{1}{1+3^3} = 0.038 \).

Apply the Trapezoidal Rule Formula

The trapezoidal rule formula for \( n = 6 \) is given by:\[ T_n = \frac{\Delta x}{2} \left( f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + 2f(x_5) + f(x_6) \right) \]Substitute the values from Step 3:\[ T_6 = \frac{0.5}{2} \left( 1 + 2(0.941) + 2(0.5) + 2(0.228) + 2(0.111) + 2(0.060) + 0.038 \right) \]\[ T_6 = 0.25 \left( 1 + 1.882 + 1.0 + 0.456 + 0.222 + 0.12 + 0.038 \right) \]\[ T_6 = 0.25 \times 4.718 = 1.1795 \]

Approximate the Integral Value

The approximate value of the integral \( \int_{0}^{3} \frac{1}{1+x^{3}} dx \) using the trapezoidal rule with \( n=6 \) is 1.180 (rounded to three decimal places).

Key concepts

Numerical Integration

Numerical integration is a valuable technique in calculus, especially useful when an integral cannot be solved analytically. Instead of finding the exact integral, numerical methods provide an approximate result, which can be sufficient for many practical applications. This is particularly beneficial for complex functions that involve roots, logarithms, or involve combinations where standard methods fall short. By dividing the integral into smaller, more manageable sections, numerical integration allows calculations over these smaller pieces to collectively represent the entire area under the curve. For our particular exercise, the trapezoidal rule is used to approximate the integral. It breaks down the integration interval from 0 to 3 into smaller segments using subintervals. Each segment is assumed to form a trapezoid, and by calculating the areas of these trapezoids, we arrive at an approximate total integral value.

Approximation

Approximation in mathematical terms involves estimating a value using various methods, which in this exercise, is achieved through the trapezoidal rule. Approximations are critical in scenarios where exact solutions are either impossible or impractical to obtain, such as evaluating complex real-world models. The beauty of approximation is that it provides a pathway to still get meaningful results efficiently and accurately enough for many needs. In the case of the trapezoidal rule, we use endpoints within each subinterval to form trapezoids. These trapezoids create an area that closes in on the true area under the graph of the function. This method finds its strength in simplicity, providing a quick yet reasonably accurate estimate.

Calculus Problem Solving

Calculus problem solving leverages principles such as differentiation and integration to tackle a wide range of challenges. At its core, it's about understanding changes, areas, and quantities. For many students, it can be overwhelming, but breaking the problems into smaller parts makes them more approachable. The trapezoidal rule, as a problem-solving tool in calculus, serves as an entry point to tackling integration problems that don't have straightforward solutions. By using this rule, we effectively convert a continuous function into a series of linear approximations. This allows us to sidestep complicated algebraic manipulations and focus on the numerical relationships between known points, easing the process of finding solutions to otherwise intractable problems.