Question

do the following with the given information.
\\(\int_{0}^{1}13\cos(x^{2})dx\\)
(a) find the approximations (t_{8}) and (m_{8}) for the given integral. (round your answer to six decimal places.)
(t_{8}=11.730327)
(m_{8}=) enter a number.
(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 fu)
(vert e_{t}vertleq0.1015625)
(vert e_{m}vertleq0.0507813)
(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
(ngeq255) for (t_{n})
(ngeq181) for (m_{n})
resources

Explanation:

Step1: Recall Mid - point rule formula

The mid - point rule $M_n$ for $\int_{a}^{b}f(x)dx$ is given by $M_n=\Delta x\sum_{i = 1}^{n}f(\bar{x}_i)$, where $\Delta x=\frac{b - a}{n}$ and $\bar{x}_i=a+(i-\frac{1}{2})\Delta x$. Here, $a = 0$, $b = 1$, $n = 8$, $\Delta x=\frac{1-0}{8}=0.125$, and $f(x)=13\cos(x^{2})$.

Step2: Calculate mid - points

The mid - points $\bar{x}_i$ for $i = 1,2,\cdots,8$ are: $\bar{x}_1=0.0625$, $\bar{x}_2=0.1875$, $\bar{x}_3=0.3125$, $\bar{x}_4=0.4375$, $\bar{x}_5=0.5625$, $\bar{x}_6=0.6875$, $\bar{x}_7=0.8125$, $\bar{x}_8=0.9375$.

Step3: Evaluate $f(\bar{x}_i)$

$f(\bar{x}_1)=13\cos((0.0625)^{2})\approx12.9997$, $f(\bar{x}_2)=13\cos((0.1875)^{2})\approx12.9543$, $f(\bar{x}_3)=13\cos((0.3125)^{2})\approx12.7737$, $f(\bar{x}_4)=13\cos((0.4375)^{2})\approx12.4407$, $f(\bar{x}_5)=13\cos((0.5625)^{2})\approx11.9497$, $f(\bar{x}_6)=13\cos((0.6875)^{2})\approx11.2961$, $f(\bar{x}_7)=13\cos((0.8125)^{2})\approx10.4873$, $f(\bar{x}_8)=13\cos((0.9375)^{2})\approx9.5314$.

Step4: Compute $M_8$

$M_8=\Delta x\sum_{i = 1}^{8}f(\bar{x}_i)=0.125\times(12.9997 + 12.9543+12.7737+12.4407+11.9497+11.2961+10.4873+9.5314)\approx11.863282$.

Answer:

$11.863282$