Question 2(d) Merit

Expand-then-integrate, a lower limit recovered from a definite integral, a separable equation in y², an area between a curve and a horizontal line, and a physical integral left entirely in symbols.

The graph shows part of y = sin² x together with the line y = 1, which touches the curve at x = −π/2 and x = π/2. The region between them is shaded.

Find the shaded area enclosed between y = sin² x, y = 1, x = −π/2 and x = π/2.

The 2024 paper, page 8 ↗ Marking schedule, page 3 ↗

Work it out

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

1. 1 Which is on top, and between which limits?

   The line y = 1 is on top, and sin² x underneath, from −π/2 to π/2.
   Area = int_−π/2^π/2 (1 − sin² x)dx

2. 2 sin² x has no standard integral. Which double-angle formula gets rid of the square?

   sin² x = 1/2 − (1/2) cos 2x, so 1 − sin² x = 1/2 + (1/2) cos 2x.

3. 3 Integrate and put in the limits.

   Bigl[(1/2) x + (1/4) sin 2xBigr]_−π/2^π/2 = (π/4 + 0) − (−π/4 + 0) = π/2

What the marker wanted

• Achieved Reaching Bigl[(1/2) x + (1/4) sin 2xBigr] with its limits.
• Merit The area, π/2, from correct integration. Any sensible rounding is accepted, and units are not needed.

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

How did it go?

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