Question

estimate $int_{0}^{1}cos(x^{2})dx$ using (a) the trapezoidal rule and (b) the midpoint rule, each with $n = 4$. give each answer correct to five decimal places.
(a) $t_{4}=$
(b) $m_{4}=$

Explanation:

Step1: Calculate $\Delta x$

We have $a = 0$, $b = 1$ and $n=4$. The formula for $\Delta x=\frac{b - a}{n}$, so $\Delta x=\frac{1 - 0}{4}=\frac{1}{4}=0.25$.

Step2: Define the sub - intervals and $x_i$ values for Trapezoidal Rule

The sub - intervals are $[0,0.25]$, $[0.25,0.5]$, $[0.5,0.75]$, $[0.75,1]$. The $x_i$ values are $x_0 = 0$, $x_1=0.25$, $x_2 = 0.5$, $x_3=0.75$, $x_4 = 1$.

Step3: Apply the Trapezoidal Rule formula

The Trapezoidal Rule formula is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]$.
$f(x)=\cos(x^{2})$, so $f(x_0)=\cos(0)=1$, $f(x_1)=\cos(0.25^{2})=\cos(0.0625)\approx0.99805$, $f(x_2)=\cos(0.5^{2})=\cos(0.25)\approx0.96891$, $f(x_3)=\cos(0.75^{2})=\cos(0.5625)\approx0.84592$, $f(x_4)=\cos(1^{2})=\cos(1)\approx0.54030$.
$T_4=\frac{0.25}{2}[1 + 2\times0.99805+2\times0.96891+2\times0.84592+0.54030]$
$=\frac{0.25}{2}[1 + 1.9961+1.93782+1.69184+0.54030]$
$=\frac{0.25}{2}[7.16606]$
$=0.89576$.

Step4: Define the mid - points for Midpoint Rule

The mid - points of the sub - intervals $[0,0.25]$, $[0.25,0.5]$, $[0.5,0.75]$, $[0.75,1]$ are $x_1^*=0.125$, $x_2^*=0.375$, $x_3^*=0.625$, $x_4^*=0.875$.

Step5: Apply the Midpoint Rule formula

The Midpoint Rule formula is $M_n=\Delta x[f(x_1^*)+f(x_2^*)+\cdots+f(x_n^*)]$.
$f(x_1^*)=\cos(0.125^{2})=\cos(0.015625)\approx0.99988$, $f(x_2^*)=\cos(0.375^{2})=\cos(0.140625)\approx0.99027$, $f(x_3^*)=\cos(0.625^{2})=\cos(0.390625)\approx0.92467$, $f(x_4^*)=\cos(0.875^{2})=\cos(0.765625)\approx0.72035$.
$M_4=0.25[0.99988 + 0.99027+0.92467+0.72035]$
$=0.25[3.63517]$
$=0.90879$.

Answer:

(a) $T_4 = 0.89576$
(b) $M_4=0.90879$