Question 18
A random variable $X$ has a probability density function given by
$$f(x)=\begin{cases}k\sin^{-1}(x), & 0\le x\le1\\0, & \text{otherwise}\end{cases}$$
where $k$ is a positive constant.
Determine the value of $k$.
The area under the curve has to be exactly 1
Watch how a rectangle lets you find the area without integrating $\sin^{-1}(x)$ at all. You will see why the answer involves $\frac{\pi}{2}-1$.
Area 2 is found by looking at the curve sideways, as $x=\sin(y)$, which is why you can integrate it so easily.
Change k until the total probability is 1
Move the slider to stretch the curve. You need the shaded area to be exactly 1 for $f$ to be a probability density function.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The marking guide has two methods, and they share the first mark.
For any pdf, the total area under the curve is 1. The function is zero outside $0\le x\le1$, so you only need to integrate over that interval.
$$k\int_{0}^{1}\sin^{-1}(x)\,dx=1\quad\ldots(1)$$
You can also write it as $\int_{0}^{1}\sin^{-1}(x)\,dx=\frac{1}{k}$.
Marker: correctly uses a suitable pdf property
$\sin^{-1}(x)$ is not on the formula sheet, so you use integration by parts. Treat the integrand as a product with the constant $k$, so you differentiate the inverse sine and integrate $k$.
$$u=\sin^{-1}(x)\Rightarrow\frac{du}{dx}=\frac{1}{\sqrt{1-x^{2}}}\qquad\frac{dv}{dx}=k\Rightarrow v=kx$$
This mark can be implied by your later working.
Marker: correctly determines both results required to progress integration by parts
Now you apply the integration by parts rule.
$$\int k\sin^{-1}(x)\,dx=kx\sin^{-1}(x)-\int\frac{kx}{\sqrt{1-x^{2}}}\,dx$$
Follow-through marks are allowed here if you made a slip in step 2.
Marker: uses integration by parts
The new integral has the derivative of $1-x^{2}$ sitting on top, up to a constant, so you substitute.
$$u=1-x^{2},\qquad\frac{du}{dx}=-2x$$
$$\int\frac{kx}{\sqrt{1-x^{2}}}\,dx=-\frac{k}{2}\int u^{-\frac{1}{2}}\,du$$
This mark can be implied by your later working.
Marker: uses a suitable substitution method to progress the developed integral within the use of integration by parts
Integrate, swap back to $x$ and put it together with step 3.
$$-\frac{k}{2}\int u^{-\frac{1}{2}}\,du=-ku^{\frac{1}{2}}=-k\sqrt{1-x^{2}}$$
$$\int k\sin^{-1}(x)\,dx=kx\sin^{-1}(x)+k\sqrt{1-x^{2}}+c$$
Watch the signs here. The minus from step 3 and the minus from the substitution cancel, so you end up adding the square root. This mark can be implied by your later working.
Marker: determines a general result for the required integral
Now you use (1) with the limits 0 and 1.
$$\begin{aligned}k\left[x\sin^{-1}(x)+\sqrt{1-x^{2}}\right]_{0}^{1}&=1\\k\left(\left(\tfrac{\pi}{2}+0\right)-(0+1)\right)&=1\\k\left(\tfrac{\pi}{2}-1\right)&=1\end{aligned}$$
$k=\dfrac{2}{\pi-2}$
You can also leave it as $\frac{1}{\frac{\pi}{2}-1}$, because the two forms are equivalent.
Marker: determines value of $k$
Instead of integrating, you can think about areas. Your integral is just the area under $y=\sin^{-1}(x)$ from 0 to 1.
$$\text{Area 1}=\int_{0}^{1}\sin^{-1}(x)\,dx=\frac{1}{k}$$
A sketch with this area shaded also earns you the mark, and so does saying that $k\sin^{-1}(x)$ has an area of 1.
Marker: correctly represents the area between $y=\sin^{-1}(x)$ and the $x$-axis for $0\le x\le1$
Now you look at the region between the curve and the $y$-axis instead. If you write the curve as $x=\sin(y)$, you can integrate with respect to $y$.
$$\text{Area 2}=\int_{0}^{\frac{\pi}{2}}\sin(y)\,dy$$
A sketch also earns the mark here, and follow-through marks are allowed.
Marker: represents the area between $y=\sin^{-1}(x)$ and the $y$-axis for $0\le y\le\frac{\pi}{2}$
This one you can integrate straight away.
$$\text{Area 2}=\left[-\cos(y)\right]_{0}^{\frac{\pi}{2}}=-\cos\left(\tfrac{\pi}{2}\right)+\cos(0)=1$$
Marker: determines the area between $y=\sin^{-1}(x)$ and the $y$-axis for $0\le y\le\frac{\pi}{2}$
The two areas fit together to make a rectangle that is 1 wide and $\frac{\pi}{2}$ high. You can see this in the animation above.
$$\frac{1}{k}+1=1\times\frac{\pi}{2}$$
Marker: determines an equation using the two areas
Rearrange for $k$.
$$\frac{1}{k}=\frac{\pi}{2}-1=\frac{\pi-2}{2}$$
$k=\dfrac{2}{\pi-2}$
This method avoids integration by parts completely. It is super useful whenever you are asked to integrate an inverse function.
Marker: determines value of $k$
Each phrase of the question gave you something
This is not a pdf you have seen before. You have to integrate $\sin^{-1}(x)$, which is not on the formula sheet, and without a calculator you need either integration by parts with a substitution inside it or the rectangle trick.
I can see how this could trip you up, because $\sin^{-1}(x)$ does not look like a product. The trick is to treat it as $k\times\sin^{-1}(x)$, so you differentiate the inverse sine and integrate the constant. The QCAA like to hide integration by parts inside a question that looks like it is about probability.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.