Cancel whole periods before you integrate

It is given that

f(x)={tan⁡(πx/(4a)),−a<x≤a,sin⁡(πx/(2a)),a<x≤3a,f(x)=\begin{cases}\tan(\pi x/(4a)),&-a<x\le a,\\\sin(\pi x/(2a)),&a<x\le3a,\end{cases}

and f(x)=f(x−4a)f(x)=f(x-4a) for all real xx, where aa is a positive real constant.

(a) Sketch y=f(x)y=f(x) for −4a≤x≤7a-4a\le x\le7a. marks

(b) Using the sketch, write down an equation relating f(x)f(x) and f(−x)f(-x). mark

(c) Show ∫−2a4af(x)dx=(ka/π)(1+ln⁡m)\int_{-2a}^{4a}f(x)dx=(ka/\pi)(1+\ln m), determining constants k,mk,m. marks

[2023/HCI/Prelim/I/7]

Total marks: 9

Solution

(a)

Sketching a periodic function

It is given that f(x)=tan⁡(πx4a)f(x)=\tan\left(\dfrac{\pi x}{4a}\right) for −a<x≤a-a<x\le a and f(x)=sin⁡(πx2a)f(x)=\sin\left(\dfrac{\pi x}{2a}\right) for a<x≤3aa<x\le3a, with f(x)=f(x−4a)f(x)=f(x-4a) for all real xx, where a>0a>0. Sketch y=f(x)y=f(x) for −4a≤x≤7a-4a\le x\le7a.

Explore this idea

Hint

  1. Sketch the tangent piece on (−a,a](-a,a]. What are its end values?

  2. Sketch the sine piece on (a,3a](a,3a]. Does it join on continuously?

  3. Repeat the pattern every 4a4a to fill [−4a,7a][-4a,7a].

Periodic tangent-sine graph and area cancellation
Periodic tangent-sine graph and area cancellationScroll across to see the full diagram.

Answer · (a)

Period 4a4a; maxima at −3a,a,5a-3a,a,5a; minima at −a,3a,7a-a,3a,7a; zeros at −4a,−2a,0,2a,4a,6a-4a,-2a,0,2a,4a,6a.

a>0a>0; join the given tangent/sine pieces continuously.

Explore the idea

Switch on the features of the graph

Drawn with a = 1; for general a every x-value is multiplied by a.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Periodic function f
  • Maxima, minima and zeros
  • Ends of one period, x = −a and x = 3a
Ideas to try
  1. Switch on “Period 4a” and check the pattern between the two lines is the same as the pattern after 3a.
  2. Which points would you label on an exam sketch?
(b)

Symmetry

Using your sketch, write down an equation relating f(x)f(x) and f(−x)f(-x).

Hint

  1. Look at the graph about the origin. Is it symmetric about the yy-axis, or does it rotate onto itself?

Answer · (b)

f(−x)=−f(x)f(-x)=-f(x).

(c)

Integrating a periodic function

Show that ∫−2a4af(x) dx=kaπ(1+ln⁡m)\displaystyle\int_{-2a}^{4a}f(x)\,dx=\frac{ka}{\pi}(1+\ln m), where kk and mm are constants to be determined.

Hint

  1. Split [−2a,4a][-2a,4a] at −a-a, aa, 3a3a.

  2. Which pieces cancel by oddness? Which are equal by periodicity?

  3. Integrate the sine piece and the tangent piece that remain.

Answer · (c)

k=−2,m=2k=-2,m=2; integral =−2a(1+ln⁡2)/π=-2a(1+\ln2)/\pi.

Signed integral, not total area.

Takeaway

Look for cancellation first

Over a whole period, the integral of this function is 0. Over [−2a,4a][-2a,4a], which is one and a half periods, only half a period survives. Spotting that cuts the work to two short integrals.