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 6.461945× enter a number. (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 good job. $m_{4}$ is an underestimate $m_{4}$ is an overestimate amazing job! what can you conclude about the true value of the integral? (round your answers to six decimal places.) < $int_{0}^{1}7cos(x^{2})dx$ < resources read it watch it master it

Explanation:

Step1: Calculate $\Delta x$

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

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

The sub - intervals are $[0,0.25]$, $[0.25,0.5]$, $[0.5,0.75]$, $[0.75,1]$. And $x_0 = 0$, $x_1=0.25$, $x_2 = 0.5$, $x_3=0.75$, $x_4 = 1$. The function is $f(x)=7\cos(x^{2})$.

Step3: Apply the Trapezoidal Rule formula

The Trapezoidal Rule formula is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)]$.
$f(x_0)=7\cos(0^{2}) = 7$;
$f(x_1)=7\cos(0.25^{2})\approx7\cos(0.0625)\approx6.978139$;
$f(x_2)=7\cos(0.5^{2})\approx7\cos(0.25)\approx6.781052$;
$f(x_3)=7\cos(0.75^{2})\approx7\cos(0.5625)\approx6.349495$;
$f(x_4)=7\cos(1^{2})\approx3.716064$.
$T_4=\frac{0.25}{2}[7 + 2\times6.978139+2\times6.781052+2\times6.349495+3.716064]$
$=\frac{0.25}{2}(7 + 13.956278+13.562104+12.69899+3.716064)$
$=\frac{0.25}{2}(60.933436)=7.616679$.

Step4: Define 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 $M_n=\Delta x[f(x_1^*)+f(x_2^*)+f(x_3^*)+f(x_4^*)]$.
$f(x_1^*)=7\cos(0.125^{2})\approx7\cos(0.015625)\approx6.999147$;
$f(x_2^*)=7\cos(0.375^{2})\approx7\cos(0.140625)\approx6.902707$;
$f(x_3^*)=7\cos(0.625^{2})\approx7\cos(0.390625)\approx6.571779$;
$f(x_4^*)=7\cos(0.875^{2})\approx7\cos(0.765625)\approx5.919293$.
$M_4=0.25(6.999147 + 6.902707+6.571779+5.919293)$
$=0.25(26.392926)=6.598232$.

Step6: Determine bounds for the integral

Since $T_4$ is an over - estimate and $M_4$ is an under - estimate, we have $6.598232<\int_{0}^{1}7\cos(x^{2})dx<7.616679$.

Answer:

(a) $7.616679$
(b) $6.598232$
$6.598232<\int_{0}^{1}7\cos(x^{2})dx<7.616679$