In mathematics, the trapezium rule (also known as the trapezoid rule, or the trapezoidal rule is an approximate technique for calculating for the definite integral.

`int_a^bf(x)dx`

The trapezoidal rule works by approximating the region under the graph of the function *f*(*x*) as a trapezoid and calculating its area. It follows that

`int_a^bf(x)dx~~(b-a) (f(a) + f(b))/(2)` (Source: Wikipedia)

For the continuous junction has f on [a, b], the trapezoidal approximation approaches the definite integral as Dx → 0^{+}, that
is,

**Proof:**

To compare the formulas for trapezoidal approximation and the Riemann sum, we see that

For dx is the positive infinitesimal, in the extra term

(½f(x_{H})-½f(x_{0}))dx

Is the infinite small, as well we follow as

**Example 1:**

To use the trapezoidal approximation with the three trapezoids to approximate the integral

`int_0^12(x^2)/(10 + x^2)dx`

**Solution:**

The diagram is given below

In the x values of interest are

x0 = 0, x1 = 4, x2 = 8, x3 = 12

Plugging the values into the function gives

f(0) = 0, f(4) = 0.6154, f(8) = 0.8649, f(12) = 0.9351

The trapezoid approximation formula will gives

12 - 0 T(n) = ______ [0 + 2(0.6154) + 2(0.8649) + 0.9351] 2(3)= 7.7914

We can compare this with the true answer of 7.8475.

**Example 2:**

To use the Trapezoidal Approximation with the 5 trap to approximate of the integral

`int_3^13sqrt(x + 13)dx`

**Solution:**

**
Error**

In the error is approximating an integral can be found by the subtracting of the true value from the estimated value. In the graphs is show that the error is directly inked to the concavity of the integrand. Without the proof, bounds for the errors using the midpoint and the trapezoid approximations are:

B(b -
a)^{3}

|E_{M}| <

24n^{2}

B(b - a)^{3}

|E_{T}| <

12n^{2}

Where

B = max |f ''(x)|

**Example 3:**

If we want to approximate

`int_0^2e^x^2` dx using the midpoint property with an error of less than .001, we compute

f''(x)=(4+4x^{2})e^{x}

which is an increasing function on [0,2], hence has its maximum at x = 2. So

B = (2 + 4(4))e^{4} < 983

so we find

983(2^{3})

<
.001

24n^{2}

or

7864 < .001(24n^{2})

n^{2} < 327667

Taking square roots of both sides gives

n > 572

Hence if we let

n = 573

we are guaranteed to have an error less than .001.