Question

estimate $int_{0}^{1}7cos(x^{2})dx$ using the trapezoidal rule and the midpoint rule, each with $n = 4$. (round your answers to six decimal places.) (a) the trapezoidal rule (b) the midpoint rule from a graph of the integrand, decide whether your answers are underestimates or overestimates. $t_{4}$ is an underestimate $t_{4}$ is an overestimate $m_{4}$ is an underestimate $m_{4}$ is an overestimate what can you conclude about the true value of the integral? (round your answers to six decimal places.) $

Explanation:

Step1: Calculate $\Delta x$

Given $a = 0$, $b = 1$, $n=4$. Then $\Delta x=\frac{b - a}{n}=\frac{1 - 0}{4}=0.25$.

Step2: Define the function

Let $f(x)=7\cos(x^{2})$.

Step3: Apply the Trapezoidal Rule

The Trapezoidal Rule formula is $T_{n}=\frac{\Delta x}{2}[f(x_{0}) + 2f(x_{1})+2f(x_{2})+\cdots+2f(x_{n - 1})+f(x_{n})]$.
$x_{0}=0$, $x_{1}=0.25$, $x_{2}=0.5$, $x_{3}=0.75$, $x_{4}=1$.
$f(x_{0})=7\cos(0)=7$
$f(x_{1})=7\cos(0.25^{2})\approx6.996564$
$f(x_{2})=7\cos(0.5^{2})\approx6.730467$
$f(x_{3})=7\cos(0.75^{2})\approx6.022777$
$f(x_{4})=7\cos(1)\approx3.716064$
$T_{4}=\frac{0.25}{2}[7 + 2\times6.996564+2\times6.730467+2\times6.022777+3.716064]\approx5.724023$

Step4: Apply the Midpoint Rule

The Midpoint Rule formula is $M_{n}=\Delta x[f(\overline{x}_{1})+f(\overline{x}_{2})+\cdots + f(\overline{x}_{n})]$, where $\overline{x}_{i}=x_{i-\frac{1}{2}}$.
$\overline{x}_{1}=0.125$, $\overline{x}_{2}=0.375$, $\overline{x}_{3}=0.625$, $\overline{x}_{4}=0.875$
$f(\overline{x}_{1})=7\cos(0.125^{2})\approx6.999878$
$f(\overline{x}_{2})=7\cos(0.375^{2})\approx6.884227$
$f(\overline{x}_{3})=7\cos(0.625^{2})\approx6.438899$
$f(\overline{x}_{4})=7\cos(0.875^{2})\approx5.524777$
$M_{4}=0.25[6.999878+6.884227+6.438899+5.524777]\approx6.461948$

Step5: Analyze over - or under - estimates

The second - derivative of $y = 7\cos(x^{2})$ is $y''=7(- 2\sin(x^{2})-4x^{2}\cos(x^{2}))$. 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.

Step6: Determine the bounds of the integral

Since $T_{4}$ is an underestimate and $M_{4}$ is an overestimate, we have $5.724023<\int_{0}^{1}7\cos(x^{2})dx<6.461948$

Answer:

(a) $5.724023$
(b) $6.461948$
$T_{4}$ is an underestimate
$M_{4}$ is an overestimate
$5.724023<\int_{0}^{1}7\cos(x^{2})dx<6.461948$