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Specialist Mathematics · Past QCAA questions All 24 questions
2022 · Paper 1 · Technology-free

Question 19

[7 marks] Technology-free
Unit 4 · Topic 1Integration techniques Tool Kit 6.3Integration by parts Tool Kit 6.2Substitution
The question

The function $f(x)$ passes through the origin.

The gradient function of $f(x)$ is defined as

$$g(x)=e^{x}\sin^{-1}\left(e^{x}\right)$$

Determine $f(x)$.

Watch the situation first

You know the slope everywhere, and one point

Watch $f(x)$ being drawn from its gradient. The height of the blue curve is the slope of the orange curve at the same $x$.

g(x) tells you the slope of f(x) at every x sin⁻¹ needs eˣ ≤ 1, so everything happens at x ≤ 0 As x moves, the slope of f(x) follows g(x) Any curve with this slope is the same shape, shifted Only one of them passes through the origin eˣ > 1 undefined x x −4 0 g(x) = eˣ sin⁻¹(eˣ), the gradient f(x), the curve you want π2 the white tangent’s slope is the height of the blue dot f(0) = 0 you use this point to find c

Integrating gives you the whole family of curves. The origin picks out the one you want.

Then watch the method

Integration by parts, then a substitution

Watch the working build one line at a time. Orange is the part you differentiate, $u$, and blue is the part you integrate, $v$.

The gradient is g(x), so f(x) is its integral You can differentiate sin⁻¹ but not integrate it, so u = sin⁻¹(eˣ) Integration by parts leaves a simpler integral The leftover integral needs a substitution Integrate and substitute back Use the origin to find c f(x) = ∫ eˣ sin⁻¹(eˣ) dx sin⁻¹ needs eˣ ≤ 1, so the curve only exists for x ≤ 0 u = sin⁻¹(eˣ) ⇒ u′ = eˣ√1 − e²ˣ v′ = eˣ ⇒ v = eˣ mark 1 ★ f(x) = eˣ sin⁻¹(eˣ) − ∫ eˣ · eˣ√1 − e²ˣ dx = eˣ sin⁻¹(eˣ) − ∫ e²ˣ√1 − e²ˣ dx mark 2 w = 1 − e²ˣ ⇒ dw = −2e²ˣ dx −∫ e²ˣ√1 − e²ˣ dx = ½ ∫ w–½ dw use a new letter, because u is already taken mark 3 mark 4 f(x) = eˣ sin⁻¹(eˣ) + √1 − e²ˣ + c mark 5 f(0) = sin⁻¹(1) + 0 + c = π2 + c = 0 ⇒ c = −π2 f(x) = eˣ sin⁻¹(eˣ) + √1 − e²ˣ − π2 mark 6

Each line is tagged with the mark it earns. The seventh mark is for setting out all of this clearly.

Now for the mathematics

Make each decision yourself

Answer each question before you look at the marking guide. Then check your constant on the graph.

5. Check your constant on the graph

Every value of $c$ gives a curve with the right gradient. Only one of them passes through the origin.

no curve for x > 0 x y −4 −2 1 −1 −2
f(0) = π/2 + c
needs to be 0
Through the origin?
QCAA marking guide · 7 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. This follows the QCAA’s first method, integration by parts and then a substitution.

Step 1 · insight mark ★1 mark

The gradient function is the derivative, so $f(x)=\int e^{x}\sin^{-1}\left(e^{x}\right)dx$. You cannot integrate $\sin^{-1}$ directly, but you can differentiate it, so make it $u$.

$$u=\sin^{-1}\left(e^{x}\right)\ \Rightarrow\ \frac{du}{dx}=\frac{e^{x}}{\sqrt{1-e^{2x}}}\qquad\frac{dv}{dx}=e^{x}\ \Rightarrow\ v=e^{x}$$

You need the chain rule for $\frac{du}{dx}$. You can also substitute $w=e^{x}$ first and then use integration by parts on $\int\sin^{-1}(w)\,dw$. That is the QCAA’s second method, and it earns the same marks.

Marker: correctly determines the expressions for $\frac{du}{dx}$ and $v$ in preparation for the use of the integration by parts rule

Step 21 mark

Use $\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$, then tidy the new integrand.

$$\begin{aligned}f(x)&=e^{x}\sin^{-1}\left(e^{x}\right)-\int e^{x}\cdot\frac{e^{x}}{\sqrt{1-e^{2x}}}\,dx\\&=e^{x}\sin^{-1}\left(e^{x}\right)-\int\frac{e^{2x}}{\sqrt{1-e^{2x}}}\,dx\end{aligned}$$

You keep follow-through marks from here if you slipped earlier.

Marker: applies the integration by parts rule and simplifies an integrand

Step 31 mark

The leftover integral has $1-e^{2x}$ under the square root, and its derivative is already on top. That tells you which substitution to use.

$$w=1-e^{2x}\quad\Rightarrow\quad\frac{dw}{dx}=-2e^{2x}\quad\Rightarrow\quad dx=\frac{dw}{-2e^{2x}}$$

The QCAA solution uses $u$ again here. It is clearer to pick a new letter, because $u$ already means $\sin^{-1}\left(e^{x}\right)$ in your working.

Marker: determines a suitable substitution variable in preparation for using a substitution method of integration

Step 41 mark

Replace everything in the integral with $w$. The $e^{2x}$ cancels.

$$-\int\frac{e^{2x}}{\sqrt{w}}\cdot\frac{dw}{-2e^{2x}}=\frac{1}{2}\int w^{-\frac{1}{2}}\,dw$$

Marker: expresses an integrand in terms of the substitution variable

Step 51 mark

Integrate, then put $w=1-e^{2x}$ back in.

$$\frac{1}{2}\cdot2w^{\frac{1}{2}}=\sqrt{1-e^{2x}}$$

$$f(x)=e^{x}\sin^{-1}\left(e^{x}\right)+\sqrt{1-e^{2x}}+c$$

You still get this mark if you forget the $+c$, but you need it for the next step.

Marker: determines a general solution for $f(x)$

Step 61 mark

The curve passes through the origin, so $f(0)=0$. Remember that $e^{0}=1$ and $\sin^{-1}(1)=\frac{\pi}{2}$.

$$f(0)=1\cdot\sin^{-1}(1)+\sqrt{1-1}+c=\frac{\pi}{2}+c=0\quad\Rightarrow\quad c=-\frac{\pi}{2}$$

$$f(x)=e^{x}\sin^{-1}\left(e^{x}\right)+\sqrt{1-e^{2x}}-\frac{\pi}{2}$$

Marker: determines a value of the constant of integration and communicates a solution

Step 7 · communication1 mark

The last mark is for how you set out your working, up to the point where you choose the substitution. You earn it by writing the integral signs and $dx$ properly, linking each line with an equals sign, and saying what you are doing, such as “using integration by parts”.

$f(x)=e^{x}\sin^{-1}\left(e^{x}\right)+\sqrt{1-e^{2x}}-\frac{\pi}{2}$

Marker: shows logical organisation communicating key steps up to the stage where an integration method using substitution is considered

Putting it all together

Each phrase of the question gave you something

“The gradient function of $f(x)$”
You integrate $g(x)$ to get $f(x)$.
“$e^{x}\sin^{-1}\left(e^{x}\right)$”
A product with $\sin^{-1}$ in it tells you to use integration by parts, with $u=\sin^{-1}\left(e^{x}\right)$.
“passes through the origin”
You substitute $(0,\ 0)$ to find $c=-\frac{\pi}{2}$.
“Determine $f(x)$”
You finish with a full expression for $f(x)$, including the value of $c$.
What makes this complex unfamiliar

You are not told which technique to use, and one technique on its own is not enough. You need integration by parts to get rid of the $\sin^{-1}$, and then a substitution to finish the integral that is left over.

The choice I would check first is which part you called $u$. The QCAA like to set integrals where the inverse trig function has to be differentiated, not integrated, so you should make $\sin^{-1}$ your $u$ straight away.

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 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.