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Chapter Contents

• Integration
• 1. The Differential
• 2. Antiderivatives and The Indefinite Integral
• 3. The Area Under a Curve
• 4. The Definite Integral
• 5. Trapezoidal Rule
• 6. Simpson’s Rule
• Riemann Sums
• Integration Mini-lectures
• Integration Problem Solver
• Millionaire Calculus Game

• Comments, Questions?

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From the math blog...

5. The Trapezoidal Rule

by M. Bourne

Problem: Find

We put u = x² + 1 and as x = 0 → 1, u = 1 → 2

So du = 2x dx

But the question does not contain an x dx term so we cannot solve it using any of the normal integration methods.

We need to use numerical approaches. When software like Mathcad or graphics calculators perform definite integrals, they use numerical methods.

We can use one of two methods:

• Trapezoidal rule
• Simpson's Rule (in the next section: 6. Simpson's Rule)

 

The Trapezoidal Rule

We saw the basic idea in our first attempt at solving the area under the arches problem above.

Instead of using rectangles, we see that trapezoids (trapeziums) give a better approximation to the area.

Let's see this in LiveMath. Note that our approximation is much better than using rectangles.

LIVEMath

Now, the area of a trapezoid (trapezium) is given by:

So the approximate area under the curve is found by adding the area of the trapezoids. (Our trapezoids are rotated 90° so that their new base is actually the height. So h = Δx.)

Area ≈

We can simplify this to give us the Trapezoidal Rule, for n trapezoids:

[This is less calculation than the form used in the text. You may use either.]

---

To find Δx for the area from x = a to x = b, we use:

and we also need

y₀ = f(a)

y₁ = f(a + Δx)

y₂ = f(a + 2Δx)

…

y_n = f(b)

---

Note:

• We get a better approximation if we take more trapezoids [up to a limit!].
• The more trapezoids we take, we have: Δx → 0.
• We can write (if the curve is above the x-axis only between x = a and x = b):

---

Exercise: Using n = 5, approximate

 

Here, a = 0 and b = 1.

y₀ = f(a) = f(0) = = 1

y₁ = f(a + Δx) = f(0.2) =

y₂ = f(a + 2Δx) = f(0.4) =

y₃ = f(a + 3Δx) = f(0.6) =

y₄ = f(a + 4Δx) = f(0.8) =

y₅ = f(b) = f(1) =

So the area ≈

So ≈1.150