Use the error formulas to find n such that the error in the approximation of the definite integral $\int _ { 3 } ^ { 6 } \frac { 1 } { x } d x$ is less than 0.0001 using the Trapezoidal Rule.
A)43
B)44
C)42
D)40
E)41

Question

Quizplus · Accepted Answer

The error bound for the Trapezoidal Rule is given by $\frac{(b-a)^3}{12n^2} \max |f''(x)|$ on [a, b]. For the given integral, $f(x) = \frac{1}{x}$, so $f''(x) = \frac{2}{x^3}$. The interval [a, b] is [3, 6], so the maximum value of $|f''(x)|$ on this interval occurs at $x = 3$, giving $\max |f''(x)| = \frac{2}{3^3} = \frac{2}{27}$. Plugging into the error formula, we want $\frac{(6-3)^3}{12n^2} \cdot \frac{2}{27} < 0.0001$, simplifying to $\frac{27}{12n^2} \cdot \frac{2}{27} < 0.0001$, or $\frac{1}{6n^2} < 0.0001$. Solving for $n$, we find $n^2 > 1666.67$, so $n > 40.82$. Therefore, $n$ must be at least 41 to ensure the error is less than 0.0001.

Use the Error Formulas to Find N Such That the Error