Question

In Exercises 1-6, (a) use the Trapezoidal Rule with n = 4 to approximate the value of the integral. (b) Use the concavity of the function to predict whether the approximation is an overestimate or an underestimate. Finally, (c) find the integral's exact value to check your answer. $$\int_{1}^{2} \frac{1}{x} d x$$

Short answer

The Trapezoidal Rule approximation will overestimate the exact value. The exact value is \(ln(2)\), which is approximately 0.693.

Step-by-step solution

Approximate using the Trapezoidal Rule

We can use the Trapezoidal Rule to approximate the integral. The general formula for this rule is: \[T_{n}=\frac{h}{2}\left[f(x_{0})+2 f(x_{1})+2 f(x_{2})+...+2 f(x_{n-1})+f(x_{n})\right]\] Here, \(h = (b-a)/n\), and \(x_i = a + i*h\), for \(i = 0, 1, ..., n\). Given that \(a=1\), \(b=2\), and \(n=4\), we find that \(h = (2-1)/4 = 1/4 = 0.25\), and thus, the trapezoidal approximation is \[T_{4}=\frac{0.25}{2}\left[\frac{1}{1}+2*\frac{1}{1.25}+2*\frac{1}{1.5}+2*\frac{1}{1.75}+\frac{1}{2}\right]\] Calculate the values in the brackets to get the approximation.

Use concavity to predict overestimate or underestimate

We know that if \(f\) is concave up on \([a, b]\), then the Trapezoidal Rule will always overestimate the exact value of the integral over this interval, and if \(f\) is concave down, the Rule will always underestimate. The second derivative of \(f(x) = 1/x\) is \[f''(x) = 2/x^3\], which is always positive. Hence, \(f(x) = 1/x\) is concave up on \((1, 2)\), so we predict the trapezoidal approximation to overestimate the integral.

Calculate the exact value

The antiderivative of \(1/x\) is \(ln|x|\). So, by the Fundamental Theorem of Calculus, \[\int_{1}^{2} \frac{1}{x} dx = ln|2| - ln|1| = ln(2) - 0 = ln(2)\] which is roughly 0.693. Take notice of this value in comparison with your approximation.