Question 1(e) Merit to Excellence

A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.

The graph shows the curves y = 3 sec² x and y = 2 tan² x. The shaded region lies between them, with y = 3 sec² x on top, from the y-axis to the line x = 1.

Find the area of the shaded region enclosed by the two curves, x = 1, and the y-axis.

The 2024 paper, page 5 ↗ Marking schedule, page 2 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Which curve is on top, and what are the limits?

   y = 3 sec² x is on top, from the y-axis (x = 0) to x = 1.
   Area = int₀¹ (3 sec² x − 2 tan² x)dx

2. 2 tan² x has no standard integral. Which identity turns it into something that does?

   tan² x = sec² x − 1, so
   3 sec² x − 2 tan² x = 3 sec² x − 2 sec² x + 2 = sec² x + 2

3. 3 Integrate, and put in the limits. Keep the calculator in radians.

   int₀¹ (sec² x + 2)dx = [tan x + 2x]₀¹ = tan 1 + 2 = 3.557

What the marker wanted

• Achieved Either integration correct: the sec² x part or the tan² x part.
• Merit Both integrations correct.
• Excellence The area, 3.557 units², with correct integrations shown, is E8; a solution with a minor error is E7. Any sensible rounding is accepted.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.