# Integral (Trapezoidal rule and mid point rule)

• starstruck_
In summary, the conversation is about finding the integral of ∫13e^(1/x) with upper and lower bounds of 2 and 1, respectively, using the trapezoidal rule and midpoint rule. The student has attempted to find the error using the formula ∫f(x)dx = approximation + error, but is unsure how to find the integral of ∫13e^(1/x). The expert suggests looking for error bounds in a textbook, which will provide an upper bound for the error. This will not give the exact value of the error, but will provide a guaranteed maximum value.
starstruck_

## Homework Statement

find:

∫13e^(1/x)

upper bound: 2
lower bound: 1

using the trapezoidal rule and midpoint rules
estimate the errors in approximation

## Homework Equations

I've done the approximations using the trapezoidal rule and midpoint rule, but I can't figure out how to calculate the error.

this is the formula:
∫f(x)dx = approximation + error

3. The Attempt at a Solution

I need to rearrange the formula to solve for the error so:

error = ∫f(x)dx- approximation the only problem is, i have no idea how to find the integral of ∫13e^(1/x)

this is as far as i can get : 13∫e^(1/x)dx

let u = 1/x
du = -x^-2 dx where x=/= 0

uh not really sure how to work with that once I plug everything in- I don't think I've seen something like this before :/

You're supposed to estimate the error, not calculate the actual error. Look in your textbook for an expression that gives an upper bound for the error.

starstruck_ said:

## Homework Statement

find:

∫13e^(1/x)

upper bound: 2
lower bound: 1

using the trapezoidal rule and midpoint rules
estimate the errors in approximation

## Homework Equations

I've done the approximations using the trapezoidal rule and midpoint rule, but I can't figure out how to calculate the error.

this is the formula:
∫f(x)dx = approximation + error 3. The Attempt at a Solution
I need to rearrange the formula to solve for the error so:

error = ∫f(x)dx- approximationthe only problem is, i have no idea how to find the integral of ∫13e^(1/x)

this is as far as i can get : 13∫e^(1/x)dx

let u = 1/x
du = -x^-2 dx where x=/= 0

uh not really sure how to work with that once I plug everything in- I don't think I've seen something like this before :/

Every textbook that covers these topics will also have error bounds that are relatively easy to compute. These will not give the exact values of the error, but will at least give upper bounds; so if the upper bound on the error is smaller than some ##\epsilon## you know that the true (unknown) error is also guaranteed to be less than ##\epsilon##. That is always the way these things are done; it would be useless to search for an exact error, since the only way to get it would be to know the exact answer already!

## What is the trapezoidal rule?

The trapezoidal rule is a method used to approximate the value of a definite integral. It involves dividing the area under a curve into trapezoids and then calculating the sum of their areas to approximate the integral.

## How is the trapezoidal rule different from the midpoint rule?

The trapezoidal rule uses trapezoids to approximate the integral, while the midpoint rule uses rectangles. The trapezoidal rule is generally more accurate than the midpoint rule, but it requires more calculations.

## When should I use the trapezoidal rule?

The trapezoidal rule is best used when the function being integrated is smooth and does not have sharp curves or corners. It is also more accurate when the function is concave or convex.

## What is the formula for the trapezoidal rule?

The formula for the trapezoidal rule is:

ab f(x) dx ≈ (b-a)/n * (f(a)/2 + f(x1) + f(x2) + ... + f(xn-1) + f(b)/2)

Where n is the number of trapezoids used and x1, x2, ... xn-1 are the equally spaced points within the interval [a,b].

## What are the advantages of using the midpoint rule?

The midpoint rule is generally faster and easier to calculate than the trapezoidal rule. It also performs better when the function being integrated is linear or close to linear. Additionally, it can be used to approximate improper integrals more accurately than the trapezoidal rule.

• Calculus and Beyond Homework Help
Replies
4
Views
794
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
730
• Calculus and Beyond Homework Help
Replies
25
Views
670
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
14
Views
636