If $f\left(x\right)\=1\+\sin \left(\sqrt{x}\ln \left(x\+1\right)\right)$, write $\frac{h^2}{12}$ $\underset{x}{\max }$$\(\|f″\|\)$ $\left(b-a\right)$, the error bound in Table 6.7.1, as $\frac{{\left(b-a\right)}^3}{12 n^2}$$\underset{x}{\max }$$\(\|f″\|\)$ and solve the inequality $\frac{{\left(4-1\right)}^3}{12 n^2}M\le {10}^{-4}$ for $n$, where $\underset{x}{M\=\max }$$\(\|f″\|\)\doteq 0.741$. This gives $n\doteq \pm 129.1$, so the appropriate value is $n\=130$. However, for $n\=105$, the Trapezoid rule actually approximates $\mathrm{\λ}$ with an error no worse than ${10}^{-4}$.

Determine $M$, the maximum of the absolute value of the second derivative of the integrand over the interval of integration.

The numeric optimizer happily finds the "correct" maximum value and returns it as the first entry of a list. (The second member of the list is another list containing the equation that states where the maximum occurred.)

With $a\=1\,b\=4\,M\doteq 0.741$, solve the inequality $\frac{{\left(b-a\right)}^3}{12 n^2}$$\underset{x}{\max }$$\(\|f″\|\)$$\le {10}^{-4}$ for $n$.

The appropriate choice of $n$ is the first positive integer greater than $129.1$. Hence, $n\=130$ guarantees that the Trapezoid rule will approximate $\mathrm{\λ}$ with an error of no more than ${10}^{-4}$. From Example 6.7.1, take $\mathrm{\λ}$ to be the number $5.078061188$. To determine the actual value of $n$ for which the Trapezoid rule approximates $\mathrm{\λ}$ with the desired accuracy, use the ApproximateInt command as per Table 6.7.3(a).

By experiment, it is determined that $n\=105$ is the smallest value of $n$ for which the Trapezoid rule approximates $\mathrm{\λ}$ with an error no worse than ${10}^{-4}$.