Question 1(b) Achieved

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 function y = 40x(5x² − 3)³, shaded between the curve and the x-axis from x = −0.4 to the origin. The curve is above the axis there.

Find the shaded area under y = 40x(5x² − 3)³.

The 2024 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 Is the shaded region above or below the axis between x = −0.4 and 0?

   Above. For x between −0.4 and 0, x is negative and 5x² − 3 is negative, so its cube is negative too, and the product of two negatives is positive. The integral is the area.

2. 2 Which substitution? Look for an inside function whose derivative is also there.

   u = 5x² − 3, so du = 10x dx, and 40x dx = 4 du.
   int 40x(5x² − 3)³ dx = int 4u³ du = u⁴ = (5x² − 3)⁴

3. 3 Put in the limits. What is 5x² − 3 at each end?

   At 0 it is −3; at −0.4 it is 5(0.16) − 3 = −2.2.
   Bigl[(5x² − 3)⁴Bigr]_−0.4⁰ = (−3)⁴ − (−2.2)⁴ = 81 − 23.4256 = 57.5744

What the marker wanted

• Achieved The area, 57.5744, from correct integration. 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.