by M. Bourne

Interactive exploration

See an applet where you can explore Simpson’s Rule and other numerical techniques:
Riemann Sums Applet

Problem: Find

We put $displaystyle{u}={x}^{2}+{1}$u=x2+1 then $displaystyle{d}{u}={2}{x} {left.{d}{x}right.}$du=2x dx.

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

We need to use numerical approaches. (This is usually how software like Mathcad or graphics calculators perform definite integrals).

We can use one of two methods:

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

Instead of using rectangles as we did in the arches problem, we’ll use trapezoids (trapeziums) and we’ll find that it gives a better approximation to the area.

The approximate area under the curve is found by adding the area of all the trapezoids.

(Recall that we write “Δx” to mean “a small change in x“.)

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

We need “right” trapezoids (which means the parallel sides are at right angles to the base), and they are rotated 90° so that their new base is actually h, as follows, and h = Δx.

So the total area is given by:

$displaystyletext{Area}approxfrac{1}{{2}}{left({y}_{{0}}+{y}_{{1}}right)}Delta{x}+$Area≈21​(y0​+y1​)Δx+ $displaystylefrac{1}{{2}}{left({y}_{{1}}+{y}_{{2}}right)}Delta{x}+$21​(y1​+y2​)Δx+ $displaystylefrac{1}{{2}}{left({y}_{{2}}+{y}_{{3}}right)}Delta{x}+ldots$21​(y2​+y3​)Δx+…

We can simplify this to give us the Trapezoidal Rule, for $displaystyle{n}$n trapezoids:

To find $displaystyleDelta{x}$Δx for the area from $displaystyle{x}={a}$x=a to $displaystyle{x}={b}$x=b, we use:

and we also need

There is an interactive applet where you can explore the Trapezoid Rule, here:

Calculus from First Principles applet

Using $displaystyle{n}={5}$n=5, approximate the integral:

Answer

This is the situation:

I have joined each of the points at the top of the vertical segments with a straight line.
Here, $displaystyle{a}={0}$a=0 and $displaystyle{b}={1}$b=1, and the width of each trapezoid is given by:

$displaystyleDelta{x}={frac{{{b}-{a}}}{{{n}}}}={frac{{{1}-{0}}}{{{5}}}}={0.2}$Δx=nb−a​=51−0​=0.2

$displaystyle{y}_{{0}}= f{{left({a}right)}}=$y0​=f(a)= $displaystyle f{{left({0}right)}}=sqrt{{{0}^{2}+{1}}}={1}$f(0)=02+1​=1
$displaystyle{y}_{{1}}= f{{left({a}+Delta{x}right)}}=$y1​=f(a+Δx)= $displaystyle f{{left({0.2}right)}}=sqrt{{{0.2}^{2}+{1}}}={1.0198039}$f(0.2)=0.22+1​=1.0198039
$displaystyle{y}_{{2}}= f{{left({a}+{2}Delta{x}right)}}=$y2​=f(a+2Δx)= $displaystyle f{{left({0.4}right)}}=sqrt{{{0.4}^{2}+{1}}}={1.0770330}$f(0.4)=0.42+1​=1.0770330
$displaystyle{y}_{{3}}= f{{left({a}+{3}Delta{x}right)}}=$y3​=f(a+3Δx)= $displaystyle f{{left({0.6}right)}}=sqrt{{{0.6}^{2}+{1}}}={1.1661904}$f(0.6)=0.62+1​=1.1661904
$displaystyle{y}_{{4}}= f{{left({a}+{4}Delta{x}right)}}=$y4​=f(a+4Δx)= $displaystyle f{{left({0.8}right)}}=sqrt{{{0.8}^{2}+{1}}}={1.2806248}$f(0.8)=0.82+1​=1.2806248
$displaystyle{y}_{{5}}= f{{left({a}+{5}Delta{x}right)}}=$y5​=f(a+5Δx)= $displaystyle f{{left({1.0}right)}}=sqrt{{{1}^{2}+{1}}}={1.4142136}$f(1.0)=12+1​=1.4142136
So we have:
$displaystyletext{Integral}approx$Integral≈

$displaystyle={1.150}$=1.150
So $displaystyle{int_{{0}}^{{1}}}sqrt{{{x}^{2}+{1}}} {left.{d}{x}right.}approx{1.150}$∫01​x2+1​ dx≈1.150
We can see in the graph above the trapezoids are very close to the original curve, so our approximation should be close to the real value. In fact, to 3 decimal places, the integral value is 1.148.