AS 91579 · 2023 paper · a rational integral, kinematics, two bounded areas
Question 1(e) Merit to Excellence
Practises Areas under & between curves
A linear-plus-reciprocal integral, a velocity in km hr⁻¹ integrated to a distance from a point 3 km away, an area between a root and a parabola, a separable equation, and an area bounded by three different curves at once.
The shaded region is bounded by three graphs: y2 = 10 − x, y = 3x and y = −sin((π x)/10). Its left edge is y = 3x, from the origin up to where it meets the top half of the parabola. Its top edge is that half of the parabola, curving down to (10, 0). Its bottom edge is the sine curve, which dips below the x-axis between the origin and (10, 0).
Find the area of the shaded region.
| y2 | = | 10 − x |
|---|---|---|
| y | = | 3x |
| y | = | −sin((π x)/10) |
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Find the corners of the region: where does y = 3x meet the parabola, and where does the parabola meet the sine curve?
On the upper half, y = √(10 − x). With y = 3x: 9x2 = 10 − x, so 9x2 + x − 10 = 0, (9x + 10)(x − 1) = 0, and the corner is (1, 3).
The parabola and y = −sin((π x)/10) both pass through (10, 0). -
2 Split the region at the x-axis. The parts above it change their top edge at x = 1.
Above the axis: under y = 3x from 0 to 1, then under √(10 − x) from 1 to 10.
Below the axis: between the axis and the sine curve, from 0 to 10. -
3 The first piece is a triangle. The second?
Triangle: 1/2 × 1 × 3 = 1.5.
int110 (10 − x)1/2 dx = Bigl[−(2/3)(10 − x)3/2Bigr]110 = 0 + 2/3 × 27 = 18 -
4 The third piece is below the axis. What sign will its integral have?
int010 −sin((π x)/10)dx = Bigl[(10/π) cos((π x)/10)Bigr]010 = (10/π)(−1 − 1) = −20/π
Negative, because the curve is under the axis, so the area is 20/π. -
5 Add the three areas.
1.5 + 18 + 20/π = 25.866
What the marker wanted
- Achieved One of the integrations correct.
- Merit The two areas found by integration, under the parabola (18) and above the sine curve (20/π), both correct.
- Excellence The total, 25.866 units², with correct integrations, is E8; a solution with one minor error is E7. The triangle may be found by integrating instead.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.