At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To solve the integral [tex]\( \int_1^5 \frac{2 \cos (3 x)}{x} \, dx \)[/tex] using numerical methods with [tex]\( n = 8 \)[/tex] subintervals, follow these steps for each method:
### (a) Trapezoidal Rule
The trapezoidal rule approximates the integral by dividing the interval [1, 5] into [tex]\( n = 8 \)[/tex] equally spaced subintervals and summing up the areas of trapezoids under the curve. Here, the interval is divided by points [tex]\( x_i \)[/tex] from [tex]\( a = 1 \)[/tex] to [tex]\( b = 5 \)[/tex] with [tex]\( x_0, x_1, \ldots, x_8 \)[/tex]. The width of each subinterval is
[tex]\[ h = \frac{b - a}{n} = \frac{5 - 1}{8} = 0.5 \][/tex]
The formula for the trapezoidal approximation is:
[tex]\[ \int_a^b f(x) \, dx \approx \frac{h}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \][/tex]
Plugging in our function [tex]\( f(x) = \frac{2 \cos (3 x)}{x} \)[/tex] and evaluating it at the points [tex]\( x_0, x_1, \ldots, x_8 \)[/tex]:
[tex]\[ \text{Trapezoidal result} \approx -0.18916600155903196 \][/tex]
### (b) Midpoint Rule
The midpoint rule approximates the integral by dividing the interval into [tex]\( n = 8 \)[/tex] subintervals and using the midpoint of each subinterval to approximate the function:
The width of each subinterval is again [tex]\( h = 0.5 \)[/tex]. The midpoint of each subinterval is
[tex]\[ x_i = a + \left( i + \frac{1}{2} \right)h \][/tex]
for [tex]\( i = 0, 1, \ldots, n-1 \)[/tex]. The midpoint rule formula is:
[tex]\[ \int_a^b f(x) \, dx \approx h \sum_{i=0}^{n-1} f(x_i) \][/tex]
Evaluating each midpoint and the function [tex]\( f(x) \)[/tex] at these points:
[tex]\[ \text{Midpoint result} \approx -0.12401899176539463 \][/tex]
### (c) Simpson's Rule
Simpson’s rule is more accurate than the trapezoidal rule or the midpoint rule under certain conditions. It requires an even number of subintervals (which we have):
[tex]\[ \int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{i=1, \, \text{odd}}^{n-1} f(x_i) + 2 \sum_{i=2, \, \text{even}}^{n-2} f(x_i) + f(x_n) \right] \][/tex]
The subinterval width [tex]\( h \)[/tex] remains [tex]\( 0.5 \)[/tex]. Simpson's rule interpretation for our function yields:
[tex]\[ \text{Simpson's result} \approx -0.14670221666727806 \][/tex]
In summary, the results for the integral [tex]\( \int_1^5 \frac{2 \cos (3 x)}{x} \, dx \)[/tex] using [tex]\( n = 8 \)[/tex] are:
(a) Trapezoidal rule: [tex]\( -0.18916600155903196 \)[/tex]
(b) Midpoint rule: [tex]\( -0.12401899176539463 \)[/tex]
(c) Simpson's rule: [tex]\( -0.14670221666727806 \)[/tex]
### (a) Trapezoidal Rule
The trapezoidal rule approximates the integral by dividing the interval [1, 5] into [tex]\( n = 8 \)[/tex] equally spaced subintervals and summing up the areas of trapezoids under the curve. Here, the interval is divided by points [tex]\( x_i \)[/tex] from [tex]\( a = 1 \)[/tex] to [tex]\( b = 5 \)[/tex] with [tex]\( x_0, x_1, \ldots, x_8 \)[/tex]. The width of each subinterval is
[tex]\[ h = \frac{b - a}{n} = \frac{5 - 1}{8} = 0.5 \][/tex]
The formula for the trapezoidal approximation is:
[tex]\[ \int_a^b f(x) \, dx \approx \frac{h}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \][/tex]
Plugging in our function [tex]\( f(x) = \frac{2 \cos (3 x)}{x} \)[/tex] and evaluating it at the points [tex]\( x_0, x_1, \ldots, x_8 \)[/tex]:
[tex]\[ \text{Trapezoidal result} \approx -0.18916600155903196 \][/tex]
### (b) Midpoint Rule
The midpoint rule approximates the integral by dividing the interval into [tex]\( n = 8 \)[/tex] subintervals and using the midpoint of each subinterval to approximate the function:
The width of each subinterval is again [tex]\( h = 0.5 \)[/tex]. The midpoint of each subinterval is
[tex]\[ x_i = a + \left( i + \frac{1}{2} \right)h \][/tex]
for [tex]\( i = 0, 1, \ldots, n-1 \)[/tex]. The midpoint rule formula is:
[tex]\[ \int_a^b f(x) \, dx \approx h \sum_{i=0}^{n-1} f(x_i) \][/tex]
Evaluating each midpoint and the function [tex]\( f(x) \)[/tex] at these points:
[tex]\[ \text{Midpoint result} \approx -0.12401899176539463 \][/tex]
### (c) Simpson's Rule
Simpson’s rule is more accurate than the trapezoidal rule or the midpoint rule under certain conditions. It requires an even number of subintervals (which we have):
[tex]\[ \int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{i=1, \, \text{odd}}^{n-1} f(x_i) + 2 \sum_{i=2, \, \text{even}}^{n-2} f(x_i) + f(x_n) \right] \][/tex]
The subinterval width [tex]\( h \)[/tex] remains [tex]\( 0.5 \)[/tex]. Simpson's rule interpretation for our function yields:
[tex]\[ \text{Simpson's result} \approx -0.14670221666727806 \][/tex]
In summary, the results for the integral [tex]\( \int_1^5 \frac{2 \cos (3 x)}{x} \, dx \)[/tex] using [tex]\( n = 8 \)[/tex] are:
(a) Trapezoidal rule: [tex]\( -0.18916600155903196 \)[/tex]
(b) Midpoint rule: [tex]\( -0.12401899176539463 \)[/tex]
(c) Simpson's rule: [tex]\( -0.14670221666727806 \)[/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.