Question

Use the methods of (1) left Riemann sum, (2) right Riemann sum, (3) Trapezoidal Rule, (4) Parabolic Rule with \(n=8\) to approximate the definite integral. Then use the Second Fundamental Theorem of Calculus to find the exact value of each integral. $$ \int_{1}^{3} \frac{1}{x^{3}} d x $$

Short answer

The integral is approximately 0.2222.

Step-by-step solution

Define the interval and subintervals

The integral is evaluated from 1 to 3, so the interval \(a, b\) is \(1, 3\). With \(n = 8\), calculate the width of each subinterval as \(\Delta x = \frac{b-a}{n} = \frac{3-1}{8} = \frac{1}{4}\). Thus, subintervals are \[1, 1.25, 1.5, 1.75, 2, 2.25, 2.5, 2.75, 3\].

Left Riemann Sum

For the left Riemann sum, use the left endpoints of each subinterval: 1, 1.25, 1.5, 1.75, 2, 2.25, 2.5, and 2.75. Calculate the sum: \L_8 = \sum_{i=0}^{7} f(x_i) \Delta x = \\left( \frac{1}{1^3} + \frac{1}{1.25^3} + \frac{1}{1.5^3} + \frac{1}{1.75^3} + \frac{1}{2^3} + \frac{1}{2.25^3} + \frac{1}{2.5^3} + \frac{1}{2.75^3} \right) \cdot \frac{1}{4}. Approximate \(L_8 \approx 0.2319\).

Right Riemann Sum

For the right Riemann sum, use the right endpoints of each subinterval: 1.25, 1.5, 1.75, 2, 2.25, 2.5, 2.75, and 3.Calculate the sum: \R_8 = \sum_{i=1}^{8} f(x_i) \Delta x = \\left( \frac{1}{1.25^3} + \frac{1}{1.5^3} + \frac{1}{1.75^3} + \frac{1}{2^3} + \frac{1}{2.25^3} + \frac{1}{2.5^3} + \frac{1}{2.75^3} + \frac{1}{3^3} \right) \cdot \frac{1}{4}. Approximate \(R_8 \approx 0.2042\).

Trapezoidal Rule

For the Trapezoidal Rule, calculate: \T_8 = \frac{\Delta x}{2} \left(f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n)\right). Using endpoints 1 and 3 and applying it to the subintervals, \T_8 = \frac{1}{8} \left(\frac{1}{1^3} + 2\frac{1}{1.25^3} + 2\frac{1}{1.5^3} + \ldots + 2\frac{1}{2.75^3} + \frac{1}{3^3}\right).Approximate \(T_8 \approx 0.2180\).

Parabolic (Simpson's) Rule

For Simpson's Rule, ensure that n=8 (which is even) matches requirement, then apply:\S_8 = \frac{\Delta x}{3} \left(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \ldots + 4f(x_{n-1}) + f(x_n)\right). Substitute into the formula:\S_8 = \frac{1}{12} \left(\frac{1}{1^3} + 4\frac{1}{1.25^3} + 2\frac{1}{1.5^3} + \ldots + 4\frac{1}{2.75^3} + \frac{1}{3^3}\right).Approximate \(S_8 \approx 0.2189\).

Exact Integral Using the Fundamental Theorem of Calculus

Using the antiderivative of \(f(x) = \frac{1}{x^3}\), which is \(-\frac{1}{2x^2}\), evaluate:\\int_{1}^{3} \frac{1}{x^3} dx = \left[-\frac{1}{2x^2}\right]_1^3 = \\left(-\frac{1}{2(3^2)}\right) - \left(-\frac{1}{2(1^2)}\right) = \\frac{1}{2} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9}.Thus, the exact value of the integral is approximately 0.2222.

Key concepts

Definite Integral

Understanding definite integrals is essential in calculus, as they provide the foundation for calculating the area under a curve within a specified interval. In the given problem, we are asked to approximate the definite integral \[ \int_{1}^{3} \frac{1}{x^{3}} \, dx \]. This problem involves finding the integral from 1 to 3, which means we are looking for the accumulated area between the curve of \(f(x) = \frac{1}{x^3}\), the \(x\)-axis, and the vertical lines \(x = 1\) and \(x = 3\).

The definite integral has some fascinating properties: it is not only about finding areas under curves, but it also helps in understanding the accumulation of quantities, such as distance, mass, or probabilities, depending on the context. Additionally, the value of a definite integral can be negative if the curve lies below the \(x\)-axis in the interval.

Calculating a definite integral exactly isn't always straightforward, so numerical methods like Riemann sums or the Trapezoidal Rule are frequently used as approximations, as demonstrated in the problem.

Trapezoidal Rule

The Trapezoidal Rule is a numerical technique for approximating the value of a definite integral. It is based on the idea of dividing the total area under a curve into trapezoids instead of rectangles, as used in Riemann sums. Why does this matter? Because trapezoids can fit the curve better and thus give us more accurate results.

In our solution, we calculated the trapezoidal approximation for \(\int_{1}^{3} \frac{1}{x^3} \, dx\) using the Trapezoidal Rule. We worked with subintervals of width \(\Delta x = \frac{1}{4}\), and applied the formula \(T_n = \frac{\Delta x}{2} (f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n))\).

Using this rule, we reached an approximation of \(T_8 \approx 0.2180\). Despite the improvement offered by the trapezoids over simple rectangles, it's still an approximation and not the exact value. The more trapezoids we use, the closer we get to the actual value of the integral.

Simpson's Rule

Simpson's Rule provides another numerical method to approximate the value of a definite integral. This method is known to give highly accurate approximations by considering parabolas instead of simple straight lines or rectangles to assess the area under the curve.

To apply Simpson's Rule, as demonstrated in the problem \[S_8 = \frac{\Delta x}{3} (f(x_0) + 4f(x_1) + 2f(x_2) + \ldots + 4f(x_{n-1}) + f(x_n))\],ensure that \(n\) is even; here \(n = 8\) and we used the given subintervals.

This method yielded a value of \(S_8 \approx 0.2189\), which although still an approximation, tends to converge faster to the true value than the Trapezoidal Rule, particularly for smooth functions.
It's crucial to know that Simpson’s Rule uses a weighted average — weights 4 and 2 for alternate terms — to refine the approximation further. Although computationally more intensive, the payoff is often a more precise approximation in fewer steps.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differential calculus and integral calculus. The Second Fundamental Theorem, in this context, provides a practical tool for evaluating definite integrals. It states that if \(F\) is an antiderivative of \(f\) on an interval \( [a, b] \), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\].

In the given exercise, we found the exact value of the integral \(\int_{1}^{3} \frac{1}{x^3} \, dx\) by first determining that the antiderivative of \(f(x) = \frac{1}{x^3}\) is \(-\frac{1}{2x^2}\). Applying the theorem, we evaluated:\[\left[-\frac{1}{2x^2}\right]_{1}^{3} = -\frac{1}{2(3^2)} + \frac{1}{2(1^2)} \]= \(\frac{1}{2} - \frac{1}{18} = \frac{4}{9} \approx 0.2222\).

This method efficiently provides the exact value of a definite integral, showcasing the power of the Fundamental Theorem of Calculus in solving integration problems with precision.