Question 5
Determine $\displaystyle\int_a^b 2\cos(x)\,dx$, where $a=\dfrac{\pi}{3}$ and $b=\dfrac{\pi}{2}$
Not quite. This is $\sin\left(\tfrac{\pi}{2}\right)-\sin\left(\tfrac{\pi}{3}\right)$, the integral of $\cos(x)$ on its own. The 2 stays in the antiderivative, $2\sin(x)$, so the answer is twice this: $2-\sqrt{3}$. In 2024, 16.5% of Queensland students chose (A).
Not quite. This is negative, but $2\cos(x)$ is positive between $\tfrac{\pi}{3}$ and $\tfrac{\pi}{2}$, so the integral must be positive. You get this value if you drop the 2 and also subtract the bounds the wrong way round, or integrate $\cos(x)$ to $-\sin(x)$. In 2024, 12.1% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2024, 57.9% of Queensland students chose (C).
Not quite. This is negative, but $2\cos(x)$ is positive between $\tfrac{\pi}{3}$ and $\tfrac{\pi}{2}$, so the integral must be positive. You get $\sqrt{3}-2$ by subtracting the bounds the wrong way round, or by integrating $\cos(x)$ to $-\sin(x)$. The antiderivative of $\cos(x)$ is $+\sin(x)$. In 2024, 13.2% of Queensland students chose (D).
This is the question’s integral. The whole region is above the axis, so the integral is positive and equals the orange area, $2-\sqrt{3}\approx0.268$.
The bounds are equal, so there is no region and the integral is 0.
The whole region is above the axis, so the integral is positive. It equals the orange area.
The whole region is below the axis, so the integral is negative. It equals minus the blue area.
Part of the region is below the axis. That part counts as negative, so the integral is the orange area minus the blue area.
The antiderivative of $\cos(x)$ is $\sin(x)$, and the 2 comes along.
$$\int 2\cos(x)\,dx=2\sin(x)+c$$
Check: $\dfrac{d}{dx}\big(2\sin(x)\big)=2\cos(x)$.
Substitute the top bound, then subtract the bottom bound. Use the exact values $\sin\left(\tfrac{\pi}{2}\right)=1$ and $\sin\left(\tfrac{\pi}{3}\right)=\tfrac{\sqrt{3}}{2}$.
$\displaystyle \Big[2\sin(x)\Big]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$$\displaystyle =2(1)-2\left(\frac{\sqrt{3}}{2}\right)$$\displaystyle =2-\sqrt{3}$
$2\cos(x)$ is positive between $\tfrac{\pi}{3}$ and $\tfrac{\pi}{2}$, so the integral is positive. That rules out (B) and (D), which are both negative.
(A) is exactly half of $2-\sqrt{3}$. It is the integral of $\cos(x)$ without the 2.
$\displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}2\cos(x)\,dx=\Big[2\sin(x)\Big]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$$\displaystyle =2\sin\left(\tfrac{\pi}{2}\right)-2\sin\left(\tfrac{\pi}{3}\right)$
$\displaystyle =2(1)-2\left(\tfrac{\sqrt{3}}{2}\right)$$\displaystyle =2-\sqrt{3}$
Integrate, then substitute both bounds into the antiderivative and subtract, using exact values.
How Queensland students answered in 2024
From the QCAA Mathematical Methods subject report for 2024. 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 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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