Error approximation for trapezoidal rule?

Question

Have I calculated the error bound incorrectly?

Question

Use the Trapezoidal Rule error, to find the smallest reasonable integer $n$ such that $E_T \leq \frac{1}{10}$ of $$\int_{1}^{3} 2\ln(t)dt$$

My work:

\begin{align}f(x) &= 2\ln(t)\\ f’(x) &= \frac{2}{x}\\ f''(x) &= - \frac{2}{x^{2}}\end{align}

Testing the end points, should I find critical points? Not sure

\begin{align}f''(1)& = \frac{2}{9}\\\\ f''(3) &= 2 \tag{larger value therefore max}\\\\ |E_T| &< \frac{(2)(2)^{3}}{12n^{2}} < \frac{1}{10}\\\\ \frac{16}{12n^{2}} &< \frac{1}{10}\\\\ \sqrt{\frac{40}{3}} &< n\end{align}

Is this incorrect? Thank you

Answer 1

I think the question is about exact error not an estimate. The integral is $$ I = \int_1^3 f(t)\, dt =6 \ln 3 - 4 \approx 2.592 $$

Trapezoidal rule with differnt $n$ gives $$ I_1 = f(3) - f(1) \approx 2.197 $$

$$ I_2 = \frac{f(3) + 2 f(2) + f(1)}{2} \approx 2.485 $$

$$ I_3 = \frac{f(3)+2f(7/3)+2f(5/3)+f(1)}{3} \approx 2.543 $$

$I_3$ is the first close enough.