Question 13
$F(x)=\displaystyle\int\left(4e^{2x}+\sin(2x)\right)dx.$ Use integration to determine $F(x)$, if $F(0)=5$.
[3 marks]If $\dfrac{dy}{dx}=\left(\dfrac{3x^7-2x}{x^4}\right)^2$, determine $y$.
[3 marks]Tidy up first, integrate term by term, then use the data point
On the left, part a): integrate, then watch $F(0)=5$ slide the curve into place. On the right, part b): turn the square into separate powers of $x$ before you integrate.
You can only integrate one term at a time. Get every term into a form you know first, and leave the data point until last.
Find the constant, then tidy the square
In part a), slide $c$ until the curve passes through the point the question gives you. In part b), choose an expansion of the square and test it against the real derivative.
The curve passes below the given point, because $F(0)$ is less than 5. Increase $c$. Changing $c$ only moves the curve up or down. Its shape comes from the integral.
The curve passes above the given point, because $F(0)$ is more than 5. Decrease $c$.
That’s it. $2e^0=2$ and $\frac{1}{2}\cos(0)=\frac{1}{2}$, so $F(0)=\frac{3}{2}+c=5$ and $c=3\frac{1}{2}$. If you took $\cos(0)$ as 0, you would get $c=3$, which is half a unit too low.
Choose an expansion. A correct one lands exactly on the blue curve at every value of $x$.
This squares each term but leaves out the middle term. $(a-b)^2=a^2-2ab+b^2$, and here $2ab=2\left(3x^3\right)\left(2x^{-3}\right)=12$. Your dashed curve sits exactly 12 above the blue one everywhere.
That’s it. The dashed curve lies on the blue one at every test value. The middle term is a constant, because $x^3\times x^{-3}=1$, so it integrates to $-12x$.
This squares the top but not the bottom. The denominator becomes $\left(x^4\right)^2=x^8$, so you divide by $x^8$, not $x^4$. At $x=1$ the two agree only because every power of 1 is 1. They also touch near $x=0.93$, where both are 0. Test another value.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Part a) has a mark for each term and one for the constant. Part b) has two marks for the algebra and one for the integral.
Integrate one term at a time. The exponential stays as it is, and you divide by 2, the coefficient of $x$ in the power.
Equivalent forms such as $2\left(e^{2x}\right)$ are accepted.
Marker: correctly integrates the exponential term
The integral of $\sin$ is $-\cos$. Divide by the 2 inside again. Then put the terms together with $+c$.
$$\int\sin(2x)\,dx=-\frac{1}{2}\cos(2x)$$
Check the sign by differentiating: $-\frac{1}{2}\cos(2x)$ gives $\sin(2x)$.
Marker: correctly integrates the trigonometric term
Use $F(0)=5$ to find $c$. At $x=0$, $e^0=1$ and $\cos(0)=1$.
$$\begin{aligned}5&=2e^{0}-\frac{1}{2}\cos(0)+c\\5&=2-\frac{1}{2}+c\\c&=3\frac{1}{2}\end{aligned}$$
$c=3.5$ is also accepted. Follow-through applies, so if an integral in steps 1 or 2 had a slip, you can still earn this mark by finding $c$ correctly from your own $F(x)$.
Marker: determines the constant of integration
You can’t integrate a squared fraction directly. Split the fraction over $x^4$ first, then expand the square with $(a-b)^2=a^2-2ab+b^2$.
$$\begin{aligned}\frac{dy}{dx}&=\left(\frac{3x^7}{x^4}-\frac{2x}{x^4}\right)^2\\&=\left(3x^3-\frac{2}{x^3}\right)^2\\&=\left(3x^3\right)^2\\&\quad-2\times3x^3\times\frac{2}{x^3}\\&\quad+\left(\frac{2}{x^3}\right)^2\end{aligned}$$
Expanding the numerator first is also accepted: $\dfrac{9x^{14}-12x^8+4x^2}{x^8}$. The bottom is $\left(x^4\right)^2=x^8$.
Marker: correctly expands the squared bracket
Simplify each term into a power of $x$. In the middle term the powers cancel, so it is just a number.
$\frac{4}{x^6}$ is accepted, but you need $4x^{-6}$ to integrate. Follow-through applies if your expansion in step 4 had a slip.
Marker: simplifies the expanded squared bracket into separate terms
Now integrate each term: add one to the power, then divide by the new power. For $x^{-6}$ the new power is $-5$.
$$y=\int\left(9x^6-12+4x^{-6}\right)dx$$
The mark is for integrating all the terms, so don’t leave one out. Equivalent forms are accepted, such as $\dfrac{9x^7}{7}-12x-\dfrac{4}{5x^5}+c$. There is no data point in part b), so $c$ stays as it is.
Marker: determines $y$ in terms of $x$ by integrating all terms
Each phrase of the question gave you something
Part a) is two standard integrals with a linear expression inside, then one substitution. Part b) is index laws and an expansion, then the power rule three times.
If I had to guess where you might lose marks, it would be the middle term in part b). $\left(3x^3-2x^{-3}\right)^2$ is not $9x^6+4x^{-6}$. The $-12$ is worth a mark, and it becomes $-12x$ in the answer. In part a), $\cos(0)$ is 1, not 0. Taking it as 0 gives $c=3$ instead of $3\frac{1}{2}$.
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 Mathematical Methods 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.
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