Review question

# Can the trapezium rule help us find the area under $y = x\ln x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7903

## Solution

The continuous function $f$ maps the real interval $\begin{equation*} a \le x \le b \end{equation*}$

into a subset of the positive real numbers. Derive the trapezium rule for the approximate evaluation of the area under the curve $y = f(x)$ between $x=a$ and $x=b$, using $(n-1)$ points of subdivision at a distance $h = (b-a)/n$ apart.

Consider the following diagram.

In this diagram, we start with the graph of $y = f(x)$ between $x=a$ and $x=b$. This is shown in black. Our goal is to approximate the area under this curve; this is the region in blue (and purple, where it overlaps with the red region).

We then fix an integer $n$, define $h = (b-a)/n$, and divide the interval $a \le x \le b$ into $n$ equal parts (of length $h$) by marking the points at $a+h$, $a+2h$, and so on. We then draw line segments between $(a,f(a))$ and $(a+h,f(a+h))$; between $(a+h,f(a+h))$ and $(a+2h,f(a+2h))$; and so on. This is the (piecewise linear) curve in dark red.

The trapezium rule says that the area under the red curve (the region in light red; this is a sequence of trapezia) is approximately the same as the area under the original graph, with the approximation becoming more exact as $n$ increases. We will write this symbolically: \begin{align*} &\text{area of the first trapezium} + \text{area of the second trapezium} + \dotsb + \text{area of the last trapezium}\\ &= \frac{f(a) + f(a+h)}{2} h + \frac{f(a+h) + f(a+2h)}{2} h + \dotsb + \frac{f(b-h) + f(b)}{2} h \\ &= \frac{h}{2} \left( f(a) + 2f(a+h) + 2f(a+2h) + \dotsb + 2f(b-h)+ f(b) \right) \\ &\approx \int_a^b f(x) \:dx. \end{align*}
Hence obtain an approximation to the integral $\begin{equation*} I = \int_3^5 x \ln x \:dx \end{equation*}$

using the points of subdivision given in the following table:

 $x$ $3.0$ $3.5$ $4.0$ $4.5$ $5.0$ $x \ln x$ $3.296$ $4.385$ $5.545$ $6.768$ $8.047$
In the above, $a = 3$, $b = 5$, $f(x) = x \ln x$, and $h = 0.5$. Hence, \begin{align*} \int_3^5 x \ln x \:dx &\approx \frac{0.5}{2} \left( f(3) + 2f(3.5) + 2f(4) + 2f(4.5) + f(5) \right) \\ &= \frac{1}{4} \left( 3.296 + 2 \times 4.385 + 2 \times 5.545 + 2 \times 6.768 + 8.047 \right) \\ &= 11.18475\\ &= 11.2 \quad\text{to 3 s.f.} \end{align*}

We can, in fact, integrate $x \ln x$ exactly using integration by parts. We find $\int_3^5 x \ln x \:dx = 11.17421860\!\ldots$, so the above approximation is not bad.