June Revision Exercise 3

JC Mathematics

Please do check through the solutions on your own, especially for questions that we did not have chance to properly discuss during class. You may whatsapp me too if you have a burning question.

Note: You should not spend more than 180mins on the entire exercise.

June Revision Exercise 3 Q1
June Revision Exercise 3 Q2
June Revision Exercise 3 Q3
June Revision Exercise 3 Q4
June Revision Exercise 3 Q5
June Revision Exercise 3 Q6
June Revision Exercise 3 Q7
June Revision Exercise 3 Q8
June Revision Exercise 3 Q9
June Revision Exercise 3 Q10
June Revision Exercise 3 Q11

Evaluating Integrals with Modulus

JC Mathematics

Many students seem to struggle when they see an integral with modulus as they do not know where to begin with. The first thing they should note is that, we cannot evaluate an integral with modulus directly, that means, we must remove (address) the modulus first.

So let’s see how we should approach such questions, considering \int_{a}^{c} |f(x)| dx, we must first know what range of values of x for which f(x) is negative, in this case, let us assume that f(x) \le 0 for a \le x \le b, where b < c.

Then \int_a^c |f(x)| dx = -\int_a^b f(x) dx + \int_b^c f(x) dx

We break the integral up into two parts, adding a negative sign to the integral part for which f(x) \le 0. Students can relate this to reflecting f(x) about the x-axis to make it a positive area.

Students may want to reference this recent ACJC Prelim 2015 Question to see if they can do it.

Definite Integral Question #1

JC Mathematics

This is question that was tested in ACJC H2 Math Prelim P1 2015. A few of my students know how to answer it but were uncertain how to express it.

Credits: ACJC Prelims 2015
Credits: ACJC Prelims 2015

Most students were concerned with (ii) of the question.

I think the easiest way to prove this, is to first avoid writing too much and attempt to show it mathematically.

\int_{\pi /3}^{2\pi /3} |cos\frac{x}{2} cosx| dx
= \int_{\pi /3}^{pi /2} cos\frac{x}{2} cosx dx - \int_{\pi /2}^{2\pi /3} cos\frac{x}{2} cosx dx

while

|\int_{\pi /3}^{2\pi /3} cos\frac{x}{2} cosx dx|
= |\int_{\pi /3}^{pi /2} cos\frac{x}{2} cosx dx - \int_{\pi /2}^{2\pi /3} cos\frac{x}{2} cosx dx|

Thus |\int_{\pi /3}^{2\pi /3} cos\frac{x}{2} cosx dx| will be smaller in magnitude.

Students can also attempt to justify using the area under graph but they must express the answers in words carefully.