Mathematical Methods · Past QCAA questions All simple familiar questions
2021 · Paper 2 · Simple familiar

Question 11

[5 marks] Technology-active
Topic guidePremiumUnit 4 Topic 1.2 · Area under a curve
Game plansPremium1.5 Area under a curve (technology-active)0.4 Solving a trig equation exactly
The question

Consider the function $f(x)=e^x\sin(x)$, $0\le x\le2\pi$

a)

State the exact values of the $x$-intercepts of the graph of $f(x)$.

[2 marks] b)

Write an expression for the area enclosed between the graph of $f(x)$ and the $x$-axis.

[2 marks] c)

Determine the area enclosed between the graph of $f(x)$ and the $x$-axis to the nearest square unit.

[1 mark]
Watch the situation first

A small hump up, a deep dip down

Here is the graph drawn to scale. Watch where it crosses the axis, which part sits below, and what that part needs before you add it on.

Where does the graph meet the x-axis? Part a): ex is never zero, so sin(x) = 0 Part b): above the axis, then below it The part below has a negative integral, so take its absolute value Part c): add the two parts, then round to the nearest unit y x y = ex sin(x) 0 π 2π 12.07 −279.32 279.32 ex sin(x) = 0 ex ≠ 0, so sin(x) = 0 x = 0, π, 2π ∫ π 0 ex sin(x) dx ≈ 12.07 ∫ 2π π ex sin(x) dx ≈ −279.32 |−279.32| = 279.32 Area ≈ 12.07 + 279.32 ≈ 291.39 Area ≈ 291 square units

Almost all of the area is below the axis. Its integral comes out negative, so it needs an absolute value before you add on the small part above.

Now for the mathematics

Where do you split, and what does the part below need?

Part b) turns on two choices: the $x$-value where you split the integral, and whether the second integral gets an absolute value. Blue strips add to the total and grey hatched strips subtract from it. The zoom shows the small part above the axis, which is hard to see at full scale.

π 2π 0 −50−100−150 y x k Zoom on the part above the axis 0π
Adds Subtracts
The second integral
With $f(x)=e^x\sin(x)$

This is part b). The split sits on the $x$-intercept $x=\pi$, the part above the axis stays as it is, and the part below gets the absolute value. Every strip adds. The total, about 291.39, is the area for part c), which is 291 square units.

The split is in the right place, on the $x$-intercept $x=\pi$. But with no absolute value, the second integral stays at about $-279.32$ and cancels the 12.07 above the axis. A total of $-267.25$ can’t be an area. Switch on the absolute value.

With no absolute value, the two integrals add straight back up to the integral from 0 to $2\pi$, about $-267.25$, wherever you split. The part below the axis still counts as negative and cancels the part above. Switch on the absolute value, then look at where the split goes.

With the split at 0, the absolute value goes around the whole integral from 0 to $2\pi$. That fixes the sign but not the cancelling. The part above the axis is now subtracted, so you get 267.25, short of the area by $2\times12.07\approx24.14$. Move the split onto $x=\pi$.

With the split at $2\pi$, the second integral is empty, so this is just the integral from 0 to $2\pi$, about $-267.25$. The part below the axis never gets its absolute value. Move the split onto $x=\pi$.

Your split is left of the $x$-intercept at $\pi$. The second integral starts while the graph is still above the axis, so the piece from $k$ to $\pi$ sits inside the absolute value with the part below, and it gets subtracted. You get , short of the area. It even rounds to 291 here, but only because the sliver is thin. The expression is still not the area. Move the split onto $x=\pi$.

Your split is right of the $x$-intercept at $\pi$. The first integral runs past it, so the strip from $\pi$ to $k$ is below the axis but outside the absolute value, and it gets subtracted. You get It even rounds to 291 here, but only because the strip is thin. The expression is still not the area. Move the split onto $x=\pi$.

QCAA marking guide · 5 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Parts a) and b) are exact work you write down yourself. Only part c) needs the calculator.

Step 1 · part a)1 mark

An $x$-intercept is where the graph meets the $x$-axis, so $y=0$. Set the function equal to zero.

$$0=e^x\sin(x)$$

Marker: correctly generates the equation required

Step 2 · part a)1 mark

The right side is a product, so use the null factor law. The exponential factor is never zero, because $e^x$ is always positive. That leaves the sine factor.

$$\begin{aligned}&e^x\ne0\\&\sin(x)=0\end{aligned}$$

On the unit circle, sine is zero at angles of 0, $\pi$ and $2\pi$. All three are inside the domain $0\le x\le2\pi$, including both endpoints.

$x=0,\ \pi,\ 2\pi$

The question asks for exact values, so write $\pi$ rather than 3.14159. If you find the intercepts with your calculator’s graph, recognise the decimals as multiples of $\pi$. The guide accepts alternative formats for the intercepts.

Marker: correctly determines the three $x$-intercepts

Step 3 · part b)1 mark

The intercepts split the domain into two pieces. Sketch the graph on your calculator, or test one value in each piece. At $x=\frac{\pi}{2}$, $f=e^{\pi/2}\times1>0$, so the graph is above the axis from 0 to $\pi$. A part above the axis needs nothing extra.

$$\int_0^{\pi}e^x\sin(x)\,dx$$

Follow-through is allowed if your intercepts in part a) were wrong.

Marker: determines expression for integral above the $x$-axis

Step 4 · part b)1 mark

At $x=\frac{3\pi}{2}$, $f=e^{3\pi/2}\times(-1)<0$, so the graph is below the axis from $\pi$ to $2\pi$. A definite integral over a part below the axis is negative, but an area can’t be. Wrap that integral in absolute value brackets, then add the two parts.

$\displaystyle\begin{aligned}&\text{Area}\\&=\int_0^{\pi}e^x\sin(x)\,dx\\&\quad+\left|\int_{\pi}^{2\pi}e^x\sin(x)\,dx\right|\end{aligned}$

The descriptor for this mark names the absolute value brackets, so write them in. Equivalent expressions are accepted, and follow-through applies if your intercepts had a slip. A single integral of $f(x)$ from 0 to $2\pi$ won’t do: the two parts would cancel each other instead of adding.

Marker: determines expression for integral below the $x$-axis (including absolute value brackets)

Step 5 · part c)1 mark

Set your calculator to radians first, because the $x$ in $\sin(x)$ is an angle in radians here. Then use its definite integral template to evaluate each integral from part b). Wrap the second one in the calculator’s absolute value function, or simply drop its negative sign, and add.

$$\begin{aligned}&\int_0^{\pi}e^x\sin(x)\,dx\\&\quad\approx12.0703\\&\int_{\pi}^{2\pi}e^x\sin(x)\,dx\\&\quad\approx-279.3162\\&\text{Area}\\&\approx12.0703+279.3162\\&=291.3865\end{aligned}$$

Area $\approx291$ square units

The question asks for the nearest square unit, and the guide does not accept decimal solutions, so give 291, not 291.39. Follow-through is allowed for errors in your earlier working. As a check, integrating the absolute value of $f(x)$ from 0 to $2\pi$ in one go gives the same 291.3865.

Marker: determines area to the nearest unit

Putting it all together

Each phrase of the question gave you something

“$f(x)=e^x\sin(x)$, $0\le x\le2\pi$”
A product with a factor that is never zero, so only the sine factor makes intercepts. The domain gives the two ends.
“the exact values of the $x$-intercepts”
Solve $\sin(x)=0$ and leave the answers in terms of $\pi$: $x=0,\ \pi,\ 2\pi$.
“an expression for the area enclosed between the graph of $f(x)$ and the $x$-axis”
Split at $x=\pi$, keep the part above as it is, and put absolute value brackets around the part below.
“to the nearest square unit”
Evaluate on the calculator in radians, then round: 291 square units, with no decimals.
What makes this simple familiar

This is the standard area-under-a-curve routine for a graph that crosses the axis. Find the intercepts, split the integral there, take the absolute value of any part below, and let the calculator evaluate. The intercepts come from $\sin(x)=0$, which you have solved many times.

If I had to guess where you might lose marks, it would be typing one integral from 0 to $2\pi$ into the calculator. It gives about $-267.25$, and dropping the sign gives 267, which is still wrong, because the part above the axis has been subtracted, not added. The scale makes this easy to miss: the part above is so small next to the part below that you might not notice it on your calculator’s graph. Two smaller traps sit in part c). Your calculator must be in radians, and the answer must be a whole number of square units.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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