Question 17
Determine the value of $b$ given $\displaystyle\int_{a}^{b}3x^{2}\,dx=117$ and $\displaystyle\int_{a}^{b-1}3x^{2}\,dx=56$ for $b>1$.
[4 marks]Two areas that share a start
Both integrals start at the same unknown $a$. Take one area from the other and $a$ is gone, leaving a strip one unit wide with area 61.
The bigger area minus the smaller one is the strip from $b-1$ to $b$. Its area, $117-56=61$, depends on $b$ alone.
Find the strip with area 61
Two unknowns and two equations. Move $a$ first and watch the orange strip. Its area never changes. Then move $b$ until the strip has area 61.
The strip from 4 to 5 has area $125-64=61$, and $b=5$ satisfies $b>1$.
With $a=2$ as well, both integrals match the question: 117 and 56.
Now move $a$ until the integrals read 117 and 56. You will land on $a=2$, though the question only asks for $b$.
The strip from $-5$ to $-4$ also has area $-64-(-125)=61$, because $y=3x^2$ is symmetric. This is the other root, $b=-4$. It solves both equations (with $a=-\sqrt[3]{181}$), but the question says $b>1$, so reject it.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA’s sample response, which subtracts one equation from the other to remove $a$.
An antiderivative of $3x^2$ is $F(x)=x^3$. Evaluate each integral with the unknown bounds left in.
$$\begin{aligned}F(b)-F(a)&=117\\b^3-a^3&=117\quad\text{(I)}\\[6pt](b-1)^3-a^3&=56\quad\text{(II)}\end{aligned}$$
Other methods are accepted. For example, writing $F(b)-F(a)=117$ and $F(b-1)-F(a)=56$, so $F(b)-F(b-1)=61$.
Marker: correctly establishes a formula for one of the integrals
Both equations contain the same $-a^3$, so subtract (II) from (I). That leaves one equation in $b$ only:
$$\begin{aligned}b^3-(b-1)^3&=117-56\\&=61\end{aligned}$$
If you made a slip in step 1, you can still earn this mark by follow-through.
Marker: determines equation in $b$
Expand $(b-1)^3$:
$$(b-1)^3=b^3-3b^2+3b-1$$
So $b^3-\left(b^3-3b^2+3b-1\right)=61$. The $b^3$ terms cancel and a quadratic is left. Divide by 3 and factorise.
$$\begin{aligned}3b^2-3b+1&=61\\3b^2-3b-60&=0\\b^2-b-20&=0\\(b-5)(b+4)&=0\\b&=-4,\ 5\end{aligned}$$
Follow-through applies here too if your equation in $b$ came from an earlier slip.
Marker: determines values of $b$
Use the condition in the question to judge the two values. Write the reason down, because the condition is what earns this mark.
$$\begin{aligned}&\text{Given }b>1\\&\therefore\ b=5\end{aligned}$$
The question does not ask for $a$, but it makes a quick check. From (I), $a^3=125-117=8$, so $a=2$, and $\int_2^4 3x^2\,dx=64-8=56$.
Marker: evaluates the reasonableness of solutions
Each phrase of the question gave you something
Each integral is easy, but they give you two equations in two unknowns, and neither one can be solved alone. You have to spot that the shared $a^3$ cancels when you subtract, expand a cube, solve the quadratic that is left, and then use the condition to pick one answer. It is not complex unfamiliar because the integrand is a single power of $x$, the question gives you the condition that decides between the roots, and the method is the same simultaneous-equation thinking you use elsewhere.
If I had to guess where you might lose marks, it would be the simultaneous equations. The 2022 subject report advises frequent practice at solving linear and non-linear simultaneous equations, with and without technology, and this is a non-linear pair you solve by hand. Trying to find $a$ first leads nowhere. Expanding $(b-1)^3$ is the other risk, since a sign slip there changes the quadratic. And don’t stop at $b=-4,\ 5$. The last mark needs you to use $b>1$ out loud.
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 2022 QCAA Mathematical Methods 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.
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