Answer:
(a) 4.152698
(b) 3.215557
Step-by-step explanation:
(a)
[tex]\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $[/tex]
n=4, so :
Each subinterval has length :
[tex]\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3[/tex]
Therefore the subintervals consist of:
[tex][1,5], [5,9], [9,13][/tex]
Now, the midpoints of these subintervals are:
[tex]\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11[/tex]
Hence:
[tex]M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698[/tex]
(b)
[tex]\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))[/tex]
n=4, so :
Each subinterval has length :
[tex]\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3[/tex]
Therefore the subintervals consist of:
[1,5], [5,9], [9,13]
The endpoints of the subintervals consist of:
5,9
Hence:
[tex]T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557[/tex]
This question is based on the concept of trapezoidal rule and midpoint rule. Hence, the answer by trapezoidal rule is 4.152698 and by using midpoint rule is 3.215557.
Given:
[tex]\bold{\int\limits^{13}_1{cos\;x^{2} } \, dx}[/tex]
using the Trapezoidal Rule and the Midpoint Rule, each with n = 4.
(a) By mid-point rule:
Now, solve it by given equation,
[tex]\bold{\int\limts^{13}_1{cos\;x^{2} } \, dx = M_n=\sum^\infty_{n=1} \, f(m_i)\Delta x}[/tex]
Given that n=4. Each sub-interval has length,
[tex]\Delta x= \dfrac{b-a}{n} =\dfrac{13-1}{4} =\dfrac{12}{4}=3[/tex]
Thus, sub-interval consists of, [1,5], [5,9], [9,13] .
Now the mid-points of these intervals are,
[tex]\dfrac{1+5}{2} =3\\\\\dfrac{5+9}{2} =7\\\\\dfrac{9+13}{2} =11[/tex]
Therefore,
[tex]\bold{M_4=3\times (\cos(3^{2}) )+3\times (\cos(7^{2}) )+3\times (\cos(11^{2}) )\;\approx 4.152698}[/tex]
(b) By Trapezoidal rule:
[tex]\int\limts^{13}_1{cos\;x^{2} } \, dx = T_n=\dfrac{\Delta x }{2} (f(x_0))+2 (f(x_1))+2 (f(x_2))........+2 (f(x_n_-_1)+ (f(x_n))\\[/tex]
Each sub-interval has length :
[tex]\Delta x= \dfrac{b-a}{n} =\dfrac{13-1}{4} =\dfrac{12}{4}=3[/tex]
Thus, sub-interval consists of, [1,5], [5,9], [9,13].
The endpoints of the sub-intervals consist of: 5,9.
Therefore,
[tex]\bold{T_4=\dfrac{3}{2} (cos(1^{2}))+2\times (cos(5^{2}))+2\times(cos(9^{2})+2\times(cos(13^{2}) \approx 3.215557}[/tex]
Hence, the answer by trapezoidal rule is 4.152698 and by using midpoint rule is 3.215557.
For further details, please refer this link:
https://brainly.com/question/17218343
