Question

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estimate $int_{0}^{1}9cos(x^{2})dx$ using the trapezoidal rule and the midpoint rule, each with $n = 4$. (round your answers to six decimal places.)
(a) the trapezoidal rule
3 x
(b) the midpoint rule
3 x
from a graph of the integrand, decide whether your answers are underestimates or overestimates.
$t_{4}$ is an underestimate
$t_{4}$ is an overestimate
x
$m_{4}$ is an underestimate
$m_{4}$ is an overestimate
x
what can you conclude about the true value of the integral? (round your answers to six decimal places.)
3 x < $int_{0}^{1}9cos(x^{2})dx$ < 3 x

Explanation:

Step1: Define the function and parameters

Let $f(x)=9\cos(x^{2})$, $a = 0$, $b = 1$, and $n = 4$. Then $\Delta x=\frac{b - a}{n}=\frac{1-0}{4}=0.25$.

Step2: Find the sub - intervals and endpoints for Trapezoidal Rule

The sub - intervals are $[0,0.25]$, $[0.25,0.5]$, $[0.5,0.75]$, $[0.75,1]$. The endpoints 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 $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]$.
\[

$$\begin{align*} T_4&=\frac{0.25}{2}[f(0)+2f(0.25)+2f(0.5)+2f(0.75)+f(1)]\\ &= 0.125[9\cos(0)+2\times9\cos(0.25^{2})+2\times9\cos(0.5^{2})+2\times9\cos(0.75^{2})+9\cos(1)]\\ &=0.125[9 + 18\cos(0.0625)+18\cos(0.25)+18\cos(0.5625)+9\cos(1)]\\ &\approx0.125[9+18\times0.998047+18\times0.968912+18\times0.845299+9\times0.540302]\\ &=0.125[9 + 17.964846+17.440416+15.215382+4.862718]\\ &=0.125\times64.483368\\ &\approx8.060421 \end{align*}$$

\]

Step4: Find the mid - points for Midpoint Rule

The mid - points of the sub - intervals 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 $M_n=\Delta x[f(x_1^*)+f(x_2^*)+\cdots+f(x_n^*)]$.
\[

$$\begin{align*} M_4&=0.25[f(0.125)+f(0.375)+f(0.625)+f(0.875)]\\ &=0.25[9\cos(0.125^{2})+9\cos(0.375^{2})+9\cos(0.625^{2})+9\cos(0.875^{2})]\\ &=0.25\times9[\cos(0.015625)+\cos(0.140625)+\cos(0.390625)+\cos(0.765625)]\\ &\approx0.25\times9[0.999877+0.990123+0.924177+0.731688]\\ &=2.25\times3.645865\\ &\approx8.203196 \end{align*}$$

\]

Step6: Analyze underestimates and overestimates

The second - derivative of $y = f(x)=9\cos(x^{2})$ is $y'=- 18x\sin(x^{2})$ and $y''=-18\sin(x^{2})-36x^{2}\cos(x^{2})$. On the interval $[0,1]$, $y''<0$, so the function is concave down. For a concave - down function, the Trapezoidal Rule gives an underestimate and the Midpoint Rule gives an overestimate.

Step7: Determine the bounds for the integral

Since $T_4\approx8.060421$ is an underestimate and $M_4\approx8.203196$ is an overestimate, we have $8.060421<\int_{0}^{1}9\cos(x^{2})dx<8.203196$.

Answer:

(a) $8.060421$
(b) $8.203196$
$T_4$ is an underestimate
$M_4$ is an overestimate
$8.060421<\int_{0}^{1}9\cos(x^{2})dx<8.203196$