# Recitation 8

Example 1: Estimate $\int_0^1 \cos(x^2)dx$ using (a) the Trapezoidal Rule and
(b) the Midpoint Rule, each with $n = 4$. From a graph of the integrand, decide whether your answers are underestimates or overestimates. What can you conclude about the true value of the integral?

Solution 1: (a) $T_4 = \frac{1}{4\times 2}[f(0)+2f(1/4)+2f(2/4)+2f(3/4)+f(1)]\approx 0.895759$. (b) $M_4=\frac{1}{4}[f(1/8)+f(3/8)+f(5/8)+f(7/8)]\approx 0.908907$. Because $f$ is concave down on $[0,1]$, $T_4$ is an underestimate and $M_4$ is an overestimate. We can conclude that $0.895759 < \int_0^1 \cos(x^2)dx< 0.908907$.

Example 2: (a) Find the approximations $T_8$ and $M_8$ for the integral $\int_0^1 \cos(x^2)dx$. (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose $n$ so that the approximations $T_n$ and $M_n$ to the integral in part (a) are accurate to within 0.0001?

Solution 2: Part (a) is similar to (a) in the previous example. (b) $f(x)=\cos(x^2)$, $f'(x)=-2x\sin(x^2)$, $f''(x)=-2\sin(x^2)-4x^2\cos(x^2)$. For $0\leq x\leq 1$, $x^2\leq 1$ implies that $|f''(x)| \leq 2 |\sin(x^2)|+4|\cos(x^2)| \leq 6$. So far $n=8$, we take $K=6, a=0, b=1$ to get $|E_T| \leq 6\times 1^3 / (12 \times 8^2) = 0.0078125$ and $|E_M| \leq 1^3 / (24 \times 8^2) = 0.00390625$. (c) In order to have $|E_T|\leq 0.0001$, we need $6\times 1^3/(12\times n^2) \leq 10^{-4}$, which means $n^2\geq 5000$. Therefore $n=71$ works for $T_n$. For $T_M$, we need $n^2 \geq 2500$. Take $n=50$ would work.