Question

do the following with the given information.
(a) find the approximations (t_{8}) and (m_{8}) for the given integral (int_{0}^{1}13cos(x^{2})dx). (round your answer to six decimal places.)
(b) estimate the errors in the approximations (t_{8}) and (m_{8}) in part (a). (use the fact that the range of the sine and cosine functions is bounded by (pm1) to estimate the maximum error. round your answer to seven decimal places.)
(c) how large do we have to choose (n) so that the approximations (t_{n}) and (m_{n}) to the integral are accurate to within (0.0001)? (use the fact that the range of the sine and cosine functions is bounded by (pm1) to estimate the maximum error.)

Explanation:

Step1: Recall Trapezoidal and Mid - point rules formulas

The Trapezoidal Rule for $\int_{a}^{b}f(x)dx$ is $T_n=\frac{\Delta x}{2}[f(x_0) + 2\sum_{i = 1}^{n - 1}f(x_i)+f(x_n)]$ and the Mid - point Rule is $M_n=\Delta x\sum_{i = 1}^{n}f(\bar{x}_i)$, where $\Delta x=\frac{b - a}{n}$, $x_i=a + i\Delta x$ and $\bar{x}_i=\frac{x_{i-1}+x_i}{2}$. The error bounds for the Trapezoidal Rule $E_T$ and Mid - point Rule $E_M$ are given by $|E_T|\leq\frac{(b - a)^3}{12n^2}K_2$ and $|E_M|\leq\frac{(b - a)^3}{24n^2}K_2$, where $K_2=\max|f''(x)|$ on $[a,b]$. For $y = f(x)=13\cos(x^2)$, first find the second - derivative.
Let $u = x^2$, then $y = 13\cos(u)$. Using the chain - rule, $y'=-26x\sin(x^2)$. Using the product - rule $(uv)' = u'v+uv'$ where $u=-26x$ and $v = \sin(x^2)$, $y''=-26\sin(x^2)-52x^2\cos(x^2)$. Since $|\sin(x^2)|\leq1$ and $|\cos(x^2)|\leq1$ on $[0,1]$, $|y''|\leq| - 26|+| - 52|=78$, so $K_2 = 78$.

Step2: Calculate $T_n$ and $M_n$ for part (a)

Here, $a = 0$, $b = 1$, $\Delta x=\frac{b - a}{n}=\frac{1}{n}$.
$T_n=\frac{1}{2n}[13\cos(0)+2\sum_{i = 1}^{n - 1}13\cos(i^2/n^2)+13\cos(1)]$ and $M_n=\frac{1}{n}\sum_{i = 1}^{n}13\cos((\frac{(2i - 1)^2}{4n^2}))$. When $n = 5$:
$\Delta x=\frac{1}{5}=0.2$
For the Trapezoidal Rule:
$x_0 = 0,x_1 = 0.2,x_2 = 0.4,x_3 = 0.6,x_4 = 0.8,x_5 = 1$
$T_5=\frac{0.2}{2}[13\cos(0)+2(13\cos(0.04)+13\cos(0.16)+13\cos(0.36)+13\cos(0.64))+13\cos(1)]$
$T_5 = 0.1[13+2\times13(\cos(0.04)+\cos(0.16)+\cos(0.36)+\cos(0.64))+13\cos(1)]$
$\cos(0.04)\approx0.9992,\cos(0.16)\approx0.9870,\cos(0.36)\approx0.9362,\cos(0.64)\approx0.8026,\cos(1)\approx0.5403$
$T_5=0.1[13 + 26(0.9992+0.9870+0.9362+0.8026)+13\times0.5403]$
$T_5=0.1[13+26\times3.725+7.0239]$
$T_5=0.1[13 + 96.85+7.0239]=11.687390$
For the Mid - point Rule:
$\bar{x}_1 = 0.1,\bar{x}_2 = 0.3,\bar{x}_3 = 0.5,\bar{x}_4 = 0.7,\bar{x}_5 = 0.9$
$M_5=0.2[13\cos(0.01)+13\cos(0.09)+13\cos(0.25)+13\cos(0.49)+13\cos(0.81)]$
$\cos(0.01)\approx0.9999,\cos(0.09)\approx0.9960,\cos(0.25)\approx0.9689,\cos(0.49)\approx0.8776,\cos(0.81)\approx0.6816$
$M_5=0.2\times13[0.9999+0.9960+0.9689+0.8776+0.6816]$
$M_5 = 2.6[4.524]$
$M_5=11.762400$

Step3: Calculate error bounds for part (b)

For the Trapezoidal Rule error bound $|E_T|\leq\frac{(b - a)^3}{12n^2}K_2$. Substituting $a = 0$, $b = 1$, $K_2 = 78$:
$|E_T|\leq\frac{(1 - 0)^3}{12n^2}\times78=\frac{78}{12n^2}=\frac{13}{2n^2}$
For the Mid - point Rule error bound $|E_M|\leq\frac{(b - a)^3}{24n^2}K_2$. Substituting $a = 0$, $b = 1$, $K_2 = 78$:
$|E_M|\leq\frac{(1 - 0)^3}{24n^2}\times78=\frac{78}{24n^2}=\frac{13}{4n^2}$
When $n = 5$:
$|E_T|\leq\frac{13}{2\times25}=0.2600000$
$|E_M|\leq\frac{13}{4\times25}=0.1300000$

Step4: Find $n$ for part (c)

We want $|E_T|\leq0.0001$ and $|E_M|\leq0.0001$.
For the Trapezoidal Rule:
$\frac{13}{2n^2}\leq0.0001$
$n^2\geq\frac{13}{2\times0.0001}=65000$
$n\geq\sqrt{65000}\approx254.95$ so $n = 255$
For the Mid - point Rule:
$\frac{13}{4n^2}\leq0.0001$
$n^2\geq\frac{13}{4\times0.0001}=32500$
$n\geq\sqrt{32500}\approx180.28$ so $n = 181$

Answer:

(a) $T_5 = 11.687390$, $M_5 = 11.762400$
(b) $|E_T|\leq0.2600000$, $|E_M|\leq0.1300000$
(c) For $T_n$, $n = 255$; for $M_n$, $n = 181$