Question 18
The graphs of $y=\cos(x)$ and $y=\cos^{3}(x)$ are shown for $0\le x\le\pi$. The total area of the regions bounded by the two graphs and the line $x=k$, where $0<k<\pi$, is $\dfrac{5}{8}$ units2.
Determine the value of $k$.
[6 marks]What the question doesn’t tell you
Have a go at the question cold first. Then open this to see the steps the question expects you to find for yourself.
- The graphs cross inside the interval. The words never say where. Solve $\cos(x)=\cos^{3}(x)$ by factorising (not dividing by $\cos(x)$): $\cos(x)\sin^{2}(x)=0$, so $x=0,\ \tfrac{\pi}{2},\ \pi$. Then read which graph is on top on each side from the diagram.
- Which graph is on top changes at $x=\tfrac{\pi}{2}$. Area between curves is $\int(\text{upper}-\text{lower})\,dx$, so split the integral there. Integrating $\cos(x)-\cos^{3}(x)$ straight from $0$ to $k$ lets the part below the axis cancel area instead of adding it.
- $\cos(x)-\cos^{3}(x)$ simplifies with the Pythagorean identity: $\cos(x)\left(1-\cos^{2}(x)\right)$ $=\cos(x)\sin^{2}(x)$. Nothing in the question points to it.
- $\cos(x)\sin^{2}(x)$ is a reverse chain rule form with inside function $\sin(x)$. Its derivative, $\cos(x)$, is already there.
- The stem doesn’t say whether $k$ is before or after the crossing. The first lobe alone has area $\tfrac13$, which is less than $\tfrac58$, so $k$ must be past $\tfrac{\pi}{2}$.
- $\sin(k)=\tfrac12$ has two solutions in $(0,\pi)$. Only the second-quadrant one, $\tfrac{5\pi}{6}$, lies past $\tfrac{\pi}{2}$.
Sweep across both lobes until the area is 5/8
Watch the line $x=k$ sweep out from the $y$-axis. The upper graph changes at $\tfrac{\pi}{2}$, so the area comes in two pieces, and the first piece is not enough on its own.
The printed figure is not to scale. This one is, so the line stops at the true value, $k=\tfrac{5\pi}{6}\approx2.618$, where the total shaded area is exactly 5/8.
Slide $k$ until the total area is 5/8
This figure is drawn to scale. Move $k$ and watch both lobes fill. The readout snaps to exact values at the special angles. Then try the trap: integrate straight through from $0$ to $k$ without splitting at $\tfrac{\pi}{2}$, and see why that can never reach $\tfrac58$.
The total area is exactly $\tfrac58$. The first lobe gives $\tfrac13$ and the piece from $\tfrac{\pi}{2}$ to $k$ gives $\tfrac{7}{24}$. Here $\sin(k)=\tfrac12$ with $k>\tfrac{\pi}{2}$, so $k=\tfrac{5\pi}{6}\approx2.618$.
On the symmetry route, $\pi-k=\tfrac{\pi}{6}$, and the missing piece from $k$ to $\pi$ is $\tfrac{1}{24}$, the mirror image of the piece from $0$ to $\tfrac{\pi}{6}$.
Here $\sin(k)=\tfrac12$, but the total area is only $\tfrac{1}{24}$. $\tfrac{\pi}{6}$ is the first-quadrant solution of $\sin(k)=\tfrac12$, and it is rejected because $k>\tfrac{\pi}{2}$.
The graphs cross here, and the first lobe is complete, with area $\tfrac13$. That is less than $\tfrac58$, so $k$ must be past $\tfrac{\pi}{2}$. From here on $\cos^{3}(x)$ is the upper graph, so the integral splits at $\tfrac{\pi}{2}$.
The first lobe can give at most $\tfrac13$, which is less than $\tfrac58$. Move $k$ past $\tfrac{\pi}{2}$.
The total is still less than $\tfrac58$. Move $k$ to the right.
The total is more than $\tfrac58$. Move $k$ to the left.
Very close. Use the step buttons.
Integrating $\cos(x)-\cos^{3}(x)$ straight from $0$ to $k$ gives $\tfrac13\sin^{3}(k)$. Past $\tfrac{\pi}{2}$ the second lobe counts as negative and cancels area instead of adding it, so this never goes above $\tfrac13$. Setting it equal to $\tfrac58$ gives $\sin^{3}(k)=\tfrac{15}{8}$, so $\sin(k)\approx1.23311>1$: no solution. Split at $\tfrac{\pi}{2}$.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Solve $\cos(x)=\cos^{3}(x)$ by factorising.
$$\begin{aligned}\cos(x)&=\cos^{3}(x)\\\cos(x)-\cos^{3}(x)&=0\\\cos(x)\left(1-\cos^{2}(x)\right)&=0\\\cos(x)\sin^{2}(x)&=0\end{aligned}$$
$$\begin{aligned}&\cos(x)=0\ \text{ or }\ \sin(x)=0\\&x=\tfrac{\pi}{2}\ \text{ or }\ x=0,\ \pi\end{aligned}$$
Since $0\le x\le\pi$, these are the only solutions.
The area has to be split wherever the graphs cross, so find the crossings first. Factorise rather than dividing by $\cos(x)$: dividing throws away $\cos(x)=0$, which is the crossing you need most.
From the graph, $\cos(x)\ge\cos^{3}(x)$ on $\left[0,\tfrac{\pi}{2}\right]$. Set up the first region, reusing the line $\cos(x)\sin^{2}(x)$ from above.
$$\begin{aligned}A_1&=\int_{0}^{\pi/2}\left(\cos(x)-\cos^{3}(x)\right)dx\\&=\int_{0}^{\pi/2}\cos(x)\sin^{2}(x)\,dx\end{aligned}$$
Area between two curves is $\displaystyle\int(\text{upper}-\text{lower})\,dx$. $\cos(x)\sin^{2}(x)$ is a function of $\sin(x)$ times the derivative of $\sin(x)$, which you can integrate.
Marker: correctly expresses $\cos(x)-\cos^3(x)$ in an integrable form
Let $u=\sin(x)$, and change the limits to $u$-values.
$$\begin{aligned}u&=\sin(x)\\\frac{du}{dx}&=\cos(x)\\dx&=\frac{du}{\cos(x)}\end{aligned}$$
$$\begin{aligned}x=0&\Rightarrow u=\sin(0)=0\\x=\tfrac{\pi}{2}&\Rightarrow u=\sin\!\left(\tfrac{\pi}{2}\right)=1\end{aligned}$$
Change the limits to $u$-values the moment you change the variable, so you never have to go back to $x$.
Write the integral in $u$ and integrate.
$$A_1=\int_{0}^{1}u^{2}\,du=\left[\frac{u^{3}}{3}\right]_{0}^{1}$$
Marker: correctly determines the antiderivative
Evaluate $A_1$ and compare it with $\tfrac58$.
$$A_1=\frac13-0=\frac13$$
$\tfrac13<\tfrac58$, so the first region is not enough. The line $x=k$ is past the crossing: $\tfrac{\pi}{2}<k<\pi$.
$$\begin{aligned}\text{Remaining area needed}&=\frac58-\frac13\\&=\frac{15}{24}-\frac{8}{24}\\&=\frac{7}{24}\end{aligned}$$
From the graph, $\cos^{3}(x)\ge\cos(x)$ on $\left[\tfrac{\pi}{2},\pi\right]$. Set the second region equal to $\tfrac{7}{24}$.
$$\begin{aligned}\int_{\pi/2}^{k}\left(\cos^{3}(x)-\cos(x)\right)dx&=\frac{7}{24}\\-\int_{\pi/2}^{k}\cos(x)\sin^{2}(x)\,dx&=\frac{7}{24}\\-\left[\frac{\sin^{3}(x)}{3}\right]_{\pi/2}^{k}&=\frac{7}{24}\end{aligned}$$
The upper graph has changed, so the order of subtraction changes with it. $\cos^{3}(x)-\cos(x)$ is the negative of the first integrand, so reuse its antiderivative with a minus sign: $-\tfrac13\sin^{3}(x)$. Keep the limits in $x$ this time, because $k$ is unknown.
A response that integrates $\cos(x)-\cos^3(x)$ from $0$ to $k$ without splitting at $x=\tfrac{\pi}{2}$ obtains $\sin^3(k)=\tfrac{15}{8}$, which has no solution. This response gains at most the first two marks.
A sign error in the area for $\tfrac{\pi}{2}\le x\le k$, such as $\tfrac13+\tfrac13\left(\sin^3(k)-1\right)=\tfrac58$ or $\tfrac13+\tfrac13\left(1+\sin^3(k)\right)=\tfrac58$, leads to an equation with no solution for $\tfrac{\pi}{2}<k<\pi$. This response does not gain the fourth or fifth marks.
Marker: correctly identifies the upper graph on each side of $x=\tfrac{\pi}{2}$, where the graphs meet
Marker: correctly determines an equation in $\sin(k)$ using the total area
Evaluate the bracket, then solve for $\sin(k)$.
$$\begin{aligned}-\left(\frac{\sin^{3}(k)}{3}-\frac{\sin^{3}\!\left(\tfrac{\pi}{2}\right)}{3}\right)&=\frac{7}{24}\\\frac13-\frac{\sin^{3}(k)}{3}&=\frac{7}{24}\\8-8\sin^{3}(k)&=7\qquad(\times24)\\\sin^{3}(k)&=\frac18\\\sin(k)&=\frac12\\k&=\frac{\pi}{6},\ \frac{5\pi}{6}\end{aligned}$$
Write $\sin^{3}\!\left(\tfrac{\pi}{2}\right)$ in the bracket, not $\sin^{3}(1)$: the 1 was a $u$-limit in Step 2 and does not belong in an $x$-bracket. Sine is positive in the first and second quadrants, so $\sin(k)=\tfrac12$ has two solutions in $(0,\pi)$. Step 3 showed $k>\tfrac{\pi}{2}$, which rejects $\tfrac{\pi}{6}$.
Since $\tfrac{\pi}{2}<k<\pi$, reject $\tfrac{\pi}{6}$.
$$k=\frac{5\pi}{6}$$
The total area of the regions bounded by the two graphs and the line $x=k$ is $\tfrac58$ units2 when $k=\tfrac{5\pi}{6}$.
Marker: correctly determines $k=\tfrac{5\pi}{6}$, rejecting $k=\tfrac{\pi}{6}$ with reasoning
Marker: shows logical organisation, having fully attempted the question
As in Method 1, take out the common factor $\cos(x)$, then use the Pythagorean identity.
$$\begin{aligned}\cos(x)-\cos^{3}(x)&=\cos(x)\left(1-\cos^{2}(x)\right)\\&=\cos(x)\sin^{2}(x)\end{aligned}$$
Marker: correctly expresses $\cos(x)-\cos^3(x)$ in an integrable form
As in Method 1, use the reverse chain rule with $u=\sin(x)$, $du=\cos(x)\,dx$.
$$\begin{aligned}\int\cos(x)\sin^{2}(x)\,dx&=\int u^{2}\,du\\&=\frac{u^{3}}{3}+c\\&=\frac13\sin^{3}(x)+c\end{aligned}$$
Marker: correctly determines the antiderivative
As in Method 1, Step 1, the graphs meet at $x=\tfrac{\pi}{2}$. Read from the diagram which graph is on top on each side.
$$\begin{aligned}&\text{The graphs meet at }x=\tfrac{\pi}{2}\\[6pt]&0<x<\tfrac{\pi}{2}:\ \text{upper graph }y=\cos(x)\\&\tfrac{\pi}{2}<x<\pi:\ \text{upper graph }y=\cos^{3}(x)\end{aligned}$$
Now test the symmetry, with $f(x)=\cos(x)-\cos^{3}(x)$.
$$\begin{aligned}f(\pi-x)&=\cos(\pi-x)-\cos^{3}(\pi-x)\\&=-\cos(x)+\cos^{3}(x)\\&=-f(x)\end{aligned}$$
Method 2 also uses the symmetry of the picture: the second lobe is the first lobe turned upside down and reflected across $x=\tfrac{\pi}{2}$, because $\cos(\pi-x)=-\cos(x)$. So the gap between the graphs, $\left|f(x)\right|$, is symmetric about $x=\tfrac{\pi}{2}$: the gap at $\pi-x$ equals the gap at $x$.
A response that integrates $\cos(x)-\cos^3(x)$ from $0$ to $k$ without splitting at $x=\tfrac{\pi}{2}$ obtains $\sin^3(k)=\tfrac{15}{8}$, which has no solution. This response gains at most the first two marks.
Marker: correctly identifies the upper graph on each side of $x=\tfrac{\pi}{2}$, where the graphs meet
Find the area of the first lobe.
$$\begin{aligned}&\int_{0}^{\pi/2}\cos(x)\sin^{2}(x)\,dx\\&\quad=\Big[\tfrac13\sin^{3}(x)\Big]_{0}^{\pi/2}\\&\quad=\tfrac13(1)-\tfrac13(0)\\&\quad=\tfrac13\end{aligned}$$
To turn the symmetry into an integral, take any $a$ with $\tfrac{\pi}{2}\le a<\pi$. On $a\le x\le\pi$ the upper graph is $y=\cos^{3}(x)$. Substitute $u=\pi-x$.
$$\begin{aligned}&\int_{a}^{\pi}\left(\cos^{3}(x)-\cos(x)\right)dx\\&\quad=\int_{a}^{\pi}-f(x)\,dx\end{aligned}$$
$$\begin{aligned}&u=\pi-x,\quad\frac{du}{dx}=-1\quad\Rightarrow\quad dx=-du\\&x=a\Rightarrow u=\pi-a,\quad x=\pi\Rightarrow u=0\\&-f(x)=-f(\pi-u)=f(u)\quad\text{(Step 3)}\end{aligned}$$
$$\begin{aligned}\int_{a}^{\pi}-f(x)\,dx&=\int_{\pi-a}^{0}f(u)\,(-du)\\&=\int_{0}^{\pi-a}f(u)\,du\\&=\int_{0}^{\pi-a}\cos(u)\sin^{2}(u)\,du\end{aligned}$$
With $a=\tfrac{\pi}{2}$, the second lobe has the same area as the first.
$$\begin{aligned}&\int_{0}^{\pi/2}\cos(u)\sin^{2}(u)\,du=\tfrac13\\[4pt]&\text{Total for }0\le x\le\pi\\&\quad=\tfrac13+\tfrac13=2\times\tfrac13=\tfrac23\end{aligned}$$
Each lobe has the same area, so the total for $0\le x\le\pi$ is twice the first lobe. The first lobe alone is less than $\tfrac58$, so $k>\tfrac{\pi}{2}$ and the region stops short of $\pi$. The missing piece, from $k$ to $\pi$, mirrors the piece from $0$ to $\pi-k$ in the first lobe, where the integrand is just $\cos(x)\sin^{2}(x)$.
$\tfrac13<\tfrac58$, so $k>\tfrac{\pi}{2}$, which gives $0<\pi-k<\tfrac{\pi}{2}$. Take $a=k$.
$$\begin{aligned}&\text{Area for }k\le x\le\pi\\&\quad=\int_{0}^{\pi-k}\cos(x)\sin^{2}(x)\,dx\\&\quad=\Big[\tfrac13\sin^{3}(x)\Big]_{0}^{\pi-k}\\&\quad=\tfrac13\sin^{3}(\pi-k)-\tfrac13\sin^{3}(0)\\&\quad=\tfrac13\sin^{3}(\pi-k)\end{aligned}$$
The region for $0\le x\le k$ is everything from $0$ to $\pi$ except the piece from $k$ to $\pi$. Total minus the missing piece equals the given area.
$$\tfrac23-\tfrac13\sin^{3}(\pi-k)=\tfrac58$$
Marker: correctly determines an equation in $\sin(k)$ using the total area
Isolate $\sin^{3}(\pi-k)$ and take the real cube root.
$$\begin{aligned}\tfrac13\sin^{3}(\pi-k)&=\tfrac23-\tfrac58=\tfrac{1}{24}\\\sin^{3}(\pi-k)&=\tfrac18\\\sin(\pi-k)&=\tfrac12\end{aligned}$$
Because $0<\pi-k<\tfrac{\pi}{2}$, the angle $\pi-k$ is in the first quadrant, so it is the reference angle itself.
$$\begin{aligned}\pi-k&=\tfrac{\pi}{6}\\k&=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}\end{aligned}$$
The first-quadrant value $k=\tfrac{\pi}{6}$ never appears, because $k>\tfrac{\pi}{2}$ was settled first. That is the reason it is rejected.
The total area of the regions bounded by the two graphs and the line $x=k$ is $\tfrac58$ units2 when $k=\tfrac{5\pi}{6}$.
Marker: correctly determines $k=\tfrac{5\pi}{6}$, rejecting $k=\tfrac{\pi}{6}$ with reasoning
Marker: shows logical organisation, having fully attempted the question
Each phrase of the question gave you something
The engine is an inversion of area between curves: the area is given and the upper limit is the unknown. Two decisions are withheld. One is a silent reach into the Pythagorean identity. The other is working out where the graphs cross and which is on top, which decides how the integral splits and which side of $\tfrac{\pi}{2}$ the answer lies.
If I had to guess where you might lose marks…
- Not splitting at $\tfrac{\pi}{2}$. $\displaystyle\int_{0}^{k}\left(\cos(x)-\cos^{3}(x)\right)dx$ $=\tfrac13\sin^{3}(k)$, so $\tfrac13\sin^{3}(k)=\tfrac58$ and $\sin^{3}(k)=\tfrac{15}{8}$. Then $\sin(k)\approx1.23311>1$, so there is no solution. This response earns at most the first two marks. Splitting at $\tfrac{\pi}{2}$ but still integrating $\cos(x)-\cos^{3}(x)$ on the second piece (the wrong upper graph) gives $\tfrac13+\tfrac13\left(\sin^{3}(k)-1\right)=\tfrac58$, which is the same equation. That counts as a sign error, so the fourth and fifth marks are lost. The dead end is the signal: go back and find where the graphs cross.
- A sign slip in the second piece. The correct second piece is $-\tfrac13\left(\sin^{3}(k)-1\right)=\tfrac13-\tfrac13\sin^{3}(k)$. Expanding the bracket as $-\tfrac13\sin^{3}(k)-\tfrac13$ gives $\sin^{3}(k)=-\tfrac{15}{8}$, with no solution. Writing it as $\tfrac13\left(1+\sin^{3}(k)\right)$ gives $\sin(k)=-\tfrac12$, and no $k$ in $(0,\pi)$ has a negative sine. Either way the fourth and fifth marks are lost.
- Dropping the lower-limit value $\sin^{3}\!\left(\tfrac{\pi}{2}\right)=1$. The second piece becomes $-\tfrac13\sin^{3}(k)$, so $\tfrac13-\tfrac13\sin^{3}(k)=\tfrac58$ and $\sin(k)=-\tfrac{\sqrt[3]{7}}{2}\approx-0.956466$, which is impossible. If a sign is also lost and you get $\sin(k)=\tfrac{\sqrt[3]{7}}{2}\approx0.956466$, that is not an exact value. On a technology-free paper, a non-standard value like this means a slip.
- Stopping at the first-quadrant angle. $\sin(k)=\tfrac12$ gives $\tfrac{\pi}{6}$ by reflex, but the total area for $k=\tfrac{\pi}{6}$ is only $\tfrac{1}{24}$. Mark 5 needs $\tfrac{5\pi}{6}$ and the reason $\tfrac{\pi}{6}$ is rejected ($k>\tfrac{\pi}{2}$). Giving $150$ (degrees) for a value on a radian axis also loses it.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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