Question 9
Determine $\displaystyle\int_0^3\pi\sin\left(\frac{\pi}{3}x\right)dx$.
Not quite. 3 is the upper limit on its own: $-3\cos(\pi)=3$. You still subtract the lower limit, $-3\cos(0)=-3$, and $3-(-3)=6$. In 2023, 21.3% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2023, 47.1% of Queensland students chose (B).
Not quite. The curve is above the $x$-axis for the whole interval, so the integral must be positive. $-3$ is what you get from $+3\cos\left(\tfrac{\pi}{3}x\right)$ at the upper limit alone. The antiderivative needs a minus sign, and you subtract the lower limit too. In 2023, 21.0% of Queensland students chose (C).
Not quite. $-6$ comes from $3\cos\left(\tfrac{\pi}{3}x\right)$, which has the wrong sign: its derivative is $-\pi\sin\left(\tfrac{\pi}{3}x\right)$. The region is above the $x$-axis, so the integral must be positive. In 2023, 9.8% of Queensland students chose (D).
This is the whole interval from 0 to 3. $F(3)-F(0)=3-(-3)=6$, which matches the shaded area. That’s option (B).
This area is $F(1.5)-F(0)=0-(-3)=3$, the value in option (A). It only covers half the interval. Move $u$ to 3.
$F(u)-F(0)$ matches the shaded area. Move $u$ to 3 to cover the interval the question asks for.
With $u=0$ there is no region yet, so the integral is 0. Move $u$ to the right.
This gives a negative number, but the shaded region is above the $x$-axis. The sign is wrong: the derivative of $3\cos\left(\tfrac{\pi}{3}x\right)$ is $-\pi\sin\left(\tfrac{\pi}{3}x\right)$.
This gives $-6$, option (D). But the shaded region is above the $x$-axis, so the integral must be positive. The sign of this antiderivative is wrong.
For a linear expression inside sine, divide by the coefficient of $x$.
$\displaystyle \int\sin(kx)\,dx$$\displaystyle =-\frac{1}{k}\cos(kx)+c$
Here $k=\tfrac{\pi}{3}$, so $\tfrac1k=\tfrac{3}{\pi}$.
$\displaystyle \int\pi\sin\left(\frac{\pi}{3}x\right)dx$$\displaystyle =-3\cos\left(\frac{\pi}{3}x\right)+c$
Substitute the upper limit, then subtract the lower limit.
$$\begin{aligned}&\Bigl[-3\cos\left(\tfrac{\pi}{3}x\right)\Bigr]_0^3\\&=-3\cos(\pi)-\bigl(-3\cos(0)\bigr)\\&=3+3\end{aligned}$$
Sense check: from 0 to 3 the graph is one hump above the $x$-axis, so the answer is positive. That rules out (C) and (D).
The hump is symmetric about $x=1.5$, and the area up to 1.5 is 3. So the whole hump is $2\times3=6$.
$\displaystyle \int_0^3\pi\sin\left(\frac{\pi}{3}x\right)dx$$\displaystyle =\Bigl[-3\cos\left(\tfrac{\pi}{3}x\right)\Bigr]_0^3$
$\displaystyle =-3\cos(\pi)+3\cos(0)$$\displaystyle =3+3$$\displaystyle =6$
The curve is above the $x$-axis on the whole interval, so a positive answer makes sense.
How Queensland students answered in 2023
From the QCAA Mathematical Methods subject report for 2023. 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 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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