Question 8
Calculate the total enclosed area between the graph of $y=x^2-x-6$ and the $x$-axis from $x=1$ to $x=5$
Not quite. 5.33 is $\int_1^5\left(x^2-x-6\right)dx$ done in one go. The graph is below the $x$-axis from 1 to 3, so that piece comes out as $-7.33$ and cancels area. Split at $x=3$ and add the sizes. In 2024, 62.3% of Queensland students chose (A).
Not quite. 7.33 is only the piece below the axis, from 1 to 3. The graph crosses at $x=3$, and the piece from 3 to 5 adds another 12.67. In 2024, 5.5% of Queensland students chose (B).
Not quite. 12.67 is only the piece above the axis, from 3 to 5. The piece from 1 to 3 is below the axis, and it adds another 7.33. In 2024, 4.2% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2024, 27.3% of Queensland students chose (D).
One integral from 1 to 5 gives 5.33, option (A). The piece below the axis comes out negative, so it cancels area instead of adding to it.
Split where the graph crosses the axis, at $x=3$. Both pieces count as positive area, so the total is $7.33+12.67=20.00$.
The single integral subtracts the blue piece from the orange piece, so it always comes out smaller than the area.
The area is the size of the blue piece plus the size of the orange piece.
Up to here the graph stays below the axis, so the integral is negative. The area is its size.
Area questions start with the graph. Factorise to find where it crosses the $x$-axis.
$$x^2-x-6=(x-3)(x+2)$$
It crosses at $x=-2$ and $x=3$. Only $x=3$ is between 1 and 5, so split there. From 1 to 3 the graph is below the axis. From 3 to 5 it is above.
Integrate each piece separately. Technology does this quickly.
$\displaystyle \int_1^3\left(x^2-x-6\right)dx$$\displaystyle \approx-7.33$
$\displaystyle \int_3^5\left(x^2-x-6\right)dx$$\displaystyle \approx12.67$
The negative answer only tells you that piece is below the axis.
Area is never negative, so take the size of each piece and add.
$\displaystyle \text{Area}\approx7.33+12.67$$\displaystyle =20.00$
On your calculator you can also do it in one step: $\int_1^5\left|x^2-x-6\right|dx=20$. The absolute value flips the piece below the axis.
$\displaystyle \int_1^3\left(x^2-x-6\right)dx$$\displaystyle =-\frac{22}{3}$
$\displaystyle \int_3^5\left(x^2-x-6\right)dx$$\displaystyle =\frac{38}{3}$
$$\text{Area}=\frac{22}{3}+\frac{38}{3}=20.00$$
The graph crosses the $x$-axis at $x=3$, so the two pieces are integrated separately and their sizes added.
How Queensland students answered in 2024
More students chose (A) than the answer. From the QCAA Mathematical Methods subject report for 2024. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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