Question 4
Using the trapezoidal rule with an interval size of 1, the approximate value of the integral $\displaystyle\int_0^3 0.5^x\,dx$ is
Not quite. 1.25 is less than the exact value of the integral, about 1.2624. This curve bends upwards, so every trapezoid’s straight top sits above it and the trapezoidal estimate has to come out above 1.2624. Check how you weighted each height. In 2021, 11.3% of Queensland students chose (A).
Not quite. 1.26 is the exact value of the integral, $\int_0^3 0.5^x\,dx=\tfrac{0.875}{\ln 2}\approx1.2624$, which your calculator gives directly. The question asks for the trapezoidal rule estimate, which comes out a little higher. In 2021, 36.9% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2021, 46.0% of Queensland students chose (C).
Not quite. 1.88 is $1+0.5+0.25+0.125=1.875$, all four heights added once each. In the trapezoidal rule the end heights count half as much as the middle ones: $\tfrac12\bigl[1+2(0.5+0.25)+0.125\bigr]$. In 2021, 4.9% of Queensland students chose (D).
An interval size of 1 means three strips. The estimate is $\tfrac12(1.125+1.5)=1.3125$, which rounds to 1.31. That is option (C).
With the estimate is about . The question asks for an interval size of 1, which is three strips.
The curve bends upwards, so each straight top sits just above it and every estimate is more than the exact area, about 1.2624. More strips get closer.
An interval size of 1 from 0 to 3 gives three strips, with edges at 0, 1, 2 and 3. Find the height of the curve at each edge.
$0.5^0=1$$0.5^1=0.5$$0.5^2=0.25$$0.5^3=0.125$
The first and last heights count once. Each middle height is shared by two trapezoids, so it counts twice. Then multiply by $\tfrac{h}{2}$, with $h=1$.
$$\begin{aligned}&A\approx\frac{h}{2}\Bigl[f(x_0)\\&\;+2\bigl(f(x_1)+f(x_2)\bigr)\\&\;+f(x_3)\Bigr]\end{aligned}$$
$$\begin{aligned}&A\approx\frac12\Bigl[1\\&\quad+2(0.5+0.25)\\&\quad+0.125\Bigr]\end{aligned}$$
Work out the bracket, then halve it.
$\tfrac12(1+1.5+0.125)$$=\tfrac12(2.625)$$=1.3125$
Sense check: your calculator gives the exact integral as about 1.2624. The curve bends upwards, so the trapezoids overestimate, and the answer should be a little above 1.26. Only 1.31 fits.
$$\begin{aligned}A&\approx\frac12\Bigl[f(0)\\&\qquad+2\bigl(f(1)+f(2)\bigr)\\&\qquad+f(3)\Bigr]\\&=\frac12\Bigl[1+2(0.5+0.25)\\&\qquad+0.125\Bigr]\\&=1.3125\\&\approx1.31\end{aligned}$$
The interval size is $h=1$, the end heights count once and the middle heights twice, and the whole bracket is halved.
How Queensland students answered in 2021
From the QCAA Mathematical Methods subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it