Question

Estimate 1 13 cos(x2) dx 0 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

Answer 1

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

$\int\limits^{13}_1 {cos(x^2)} \, dx =M_n= \sum_{n=1}^{\infty} f(m_i)\Delta x$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

$[1,5], [5,9], [9,13]$

Now, the midpoints of these subintervals are:

$\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11$

Hence:

$M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698$

(b)

$\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

$T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557$