Walk through an example using the trapezoid rule, then try a couple of practice problems on your own.
By now you know that we can use Riemann sums to approximate the area under a function. Riemann sums use rectangles, which make for some pretty sloppy approximations. But what if we used trapezoids to approximate the area under a function instead?
Key idea: By using trapezoids (aka the "trapezoid rule") we can get more accurate approximations than by using rectangles (aka "Riemann sums").
An example of the trapezoid rule
Let's check it out by using three trapezoids to approximate the area under the function on the interval .
Here's how that looks in a diagram when we call the first trapezoid , the second trapezoid , and the third trapezoid :
Recall that the area of a trapezoid is where is the height and and are the bases.
Finding the area of
We need to think about the trapezoid as if it's lying sideways.
The height is the at the bottom of that spans to .
The first base is the value of at , which is .
The second base is the value of at , which is .
Here's how all of this looks visually:
Let's put this all together to find the area of :
Finding the area of
Let's find the height and both of the bases:
Plug in and simplify:
Find the area of
Finding the total area approximation
We find the total area by adding up the area of each of the three trapezoids:
Here's the final simplified answer:
You should pause here and walk through the algebra to make sure you understand how we got this!
Choose the expression that uses four trapezoids to approximate the area under the function on the interval .
Choose the expression that uses three trapezoids to approximate the area under the function on the interval .
Want to join the conversation?
- In the answers for every problem why is everything but the first and last term times 2? Why isn't everything times 2?(15 votes)
- The first and last terms are the outer bases of the trapezoids on each end of the graph, whereas the inner terms are the bases of the two trapezoids either side of the term. So when you sum the areas of all the trapezoids you can simplify by saying 2 times the inner terms, rather than adding them twice.(48 votes)
- Where is he getting the ln(x) from?(0 votes)
- The original formula for the graph was F(x) = 3*ln(x). So when evaluating any Y at a specific X, you have 3*ln of that X to give you that Y. For example, at X=2, Y=3*ln(2)(9 votes)
- Is there any formula to find the error? like the trapezoid method gives us approximate area, so can we have some solutions to find the range for this approximation?(2 votes)
- Since you eventually learn how to find the exact area under the curve I never learned it, but really taking that and then subtracting the trapeoidal sum would get you the error(5 votes)
- So I'm learning Numerical Integration. It is part of numerical integration? Is the Trapezoid Rule a derivation of Riemann's Sums?(0 votes)
- Trapezoid Rule is a form of Riemann's Summs, but it uses trapezoids not rectangles. Also, this explains why integration works, integration takes the limit as number of shapes approaches infinity. Which is the area under the curve.(8 votes)
- It looks like there might be an error in the solution for the second last problem. In the working out for T3 it shows:
T3 = (1/2)(2ln(5)+2ln(6.5))*(3/2) = (3/2)(ln(5)+3ln(6.5))
But that 3 in front of the ln(6.5) on the RHS of the equation shouldn't be there, should it?(3 votes)
- the last question I don't understand how its plus two on each interval and where that came from? also, why is it only on the first and second intervals where we add the two and not the last two?(2 votes)
- From the different Riemann sum methods (Left, Right, Midpoint, Trapezoid), would the Trapezoidal sums be the most accurate?
It seems like they would be, unless there are other methods that I'm not aware of?(1 vote)
- Instead of integrating, why do we need to learn left, right, midpoint, trapezoid sums that give us inaccurate answers?(1 vote)
- 1)You may not know the precise function.
2)You are not able to integrate everything, or you can but but the problem is not well suited for the approach.(1 vote)
- It's unclear what the objective is?
If the objective is how to find the proximate area using Calculus that's OK, but I believe there is a much simpler way, that should give you the exact area by simply building boxes or rectangles over the entire area and dividing by 2, in the same way as you would find the area of a triangle, or have I missed the point of the entire thing?(1 vote)
- Once you learn patterns in integrals with infinitismally small sigma sums you'll understand.(1 vote)
- Are the formulae for midpoint sums and trapezoidal sums same ?(0 votes)