June Revision Exercise 3 Q5

(a)
$\int_0^{\mathrm{ln}2} \frac{e^3x}{e^x+2} ~dx$

$= \int_3^4 \frac{(u-2)^2}{u} ~du$

$= \int_3^4 (u-4+\frac{4}{u}) ~du$

$= \frac{1}{2}u^2 - 4u + 4 \mathrm{ln}u \Big|_3^4$

$= 4 \mathrm{ln}\frac{4}{3} - \frac{1}{2}$

(b)
$\int_0^2 |e^x - 2| dx = \int_e^{e^{\sqrt{k}}} \frac{1}{x \mathrm{ln}x} ~dx$

$- \int_0^{\mathrm{ln}2} (e^x - 2) dx + \int_{\mathrm{ln}2}^2 (e^x - 2) ~dx = \int_e^{e^{\sqrt{k}}} \frac{\frac{1}{x}}{\mathrm{ln}x} ~dx$

$-(e^x - 2x) \Big|_0^{\mathrm{ln}2} + (e^x - 2x) \Big|_{\mathrm{ln}2}^2 = \mathrm{ln}\sqrt{k}$

$\mathrm{ln}k = 2 (e^2 + 4 \mathrm{ln}2 -7)$

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June Revision Exercise 3 Q4

(i)
$A = 2\{ \frac{1}{n} (1) + \frac{1}{n}e^{\frac{1}{4n}} + \frac{1}{n}e^{\frac{2}{4n}} + \ldots + \frac{1}{n}e^{\frac{n-1}{4n}}\}$

$= \frac{2}{n} \{e^{\frac{0}{4n}} + e^{\frac{1}{4n}} + e^{\frac{2}{4n}} + \ldots + e^{\frac{n-1}{4n}} \}$

$= \frac{2}{n}\sum_{r=0}^{n-1} e^{\frac{r}{4n}}$

(ii)
Area of R $2 \int_0^1 e^{\frac{y}{4}} ~dy = 8(e^{\frac{1}{4}}-1)$

As $n \rightarrow \infty, A \rightarrow \text{Actual Area} = 8(e^{\frac{1}{4}}-1)$

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June Revision Exercise 3 Q3

(a)(i)

Let $AB=y=\frac{40-x}{2}$

$h = \sqrt{y^2-(\frac{x}{2})^2}=2\sqrt{100-5x}$

$z = \frac{1}{2}xh=\frac{1}{2}x(2\sqrt{100-5x}=x\sqrt{100-5x}$

(a)(ii)

$\frac{dz}{dx}=\sqrt{100-5x} + x\frac{1}{2}(-5)(100-5x)^{-\frac{1}{2}}$

$\frac{dz}{dx}=\frac{100-\frac{15}{2}x}{\sqrt{100-5x}}$

$\frac{dz}{dx}=0 \Rightarrow x=\frac{40}{3}$

*Students are expected to prove that $x = \frac{40}{3}$ gives the maximum area.

(b)(i)
$x^3 + 2y^3 +3xy=k$

$3x^2+6y^2\frac{dy}{dx}+3x\frac{dy}{dx}+3y=0$

$\frac{dy}{dx}=\frac{-y-x^2}{2y^2+x} (b)(ii) Tangent parallel to$latex x\$-axis $\Rightarrow \frac{dy}{dx}=0$

$y=-x^2$

Sub $y=-x^2$ into $x^3 + 2y^3 +3xy=k$

$x^3 + 2(-x^2)^3 +3x(-x^2)=k$

$2x^6 + 2x^3 + k=0$

(b)(iii)
When the line $y=-1$ is a tangent to C,

$-1=-x^2$

$x =\pm x$

When $x = 1, k=-4$

When $x = -1, k=0$

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June Revision Exercise 3 Q2

(i)

$\frac{dx}{dt}=-\frac{3a}{t^4}$

$\frac{dy}{dt}=-\frac{a}{t^2}$

$\frac{dy}{dx}= \frac{dy}{dt} \times \frac{dt}{dx} = \frac{t^2}{3}$

When $t= \frac{1}{2}, \frac{dy}{dx}= \frac{1}{12}$

Equation of tangent: $y-2a = \frac{1}{12}(x-8a) \Rightarrow y=\frac{1}{12}x + \frac{4}{3}a$

Equation of normal: $y-2a = -12(x-8a) \Rightarrow y=-12x+98a$

(ii)
$\frac{a}{t}=\frac{1}{12}\frac{a}{t^3}+\frac{4}{3}a$

$16t^3 - 12t^2 + 1 =0$

$t=\frac{1}{2} \text{ (rej) or } t=-\frac{1}{4}$

$\text{When } t=-\frac{1}{4}, x= -64a, y=-4a$

Hence, tangent cuts curve again at $(-64a, -4a)$

(iii)
At Q, $y=0 \Rightarrow x=-16a$

At R, $y=0 \Rightarrow x=\frac{49}{6}a$

$\text{Area } = \frac{1}{2}[\frac{49}{6}a-(-16a)](2a) = \frac{145}{6}a^2$ $\text{units}^2$

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June Revision Exercise 3 Q1

(i)

$2x^2 + 8xy + 5y^2 = -3$

$4x + (8x \frac{dy}{dx} + 8y) + 10y \frac{dy}{dx}=0$

$\frac{dy}{dx} = \frac{2x+4y}{4x+5y}$

For tangents to be parallel to the $x$-axis, $\frac{dy}{dx}=0$

$\Rightarrow x=-2y$

Sub $x=-2y$ into $2x^2 + 8xy + 5y^2 = -3$

$y= \pm 1$

Thus, equations of tangents which are parallel to $x$-axis are $y= 1 \text{ or } y=-1$

(ii)

$\frac{dy}{dx}=\frac{dy}{dx}\bullet \frac{dx}{dt} = \frac{2[3+2(-1.15)]}{4(3)+5(-1.15)} = -0.488$ units per second.

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June Revision Exercise 3

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.

Question of the Day #15

This question was sent in by a student, which I think is rather interesting given 2015 A’levels P1 Q3 had something similar.

Use a definite integral to evaluate

$\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{2n^2 + k^2}{n^3 + k^3}$,

Hint: Recall that the sum of area under the graph can be found using integration.

Integration Question #2

I came across the integral a few days ago. Pretty interesting, many ways to solve this integral, be it graphically or by substitution.

$\int_a^b \sqrt{(x-a)(b-x)} dx =$?

Answer: $\pi \frac{(a-b)^2}{4}$

Hint: Consider the graph of $y = \sqrt{(x-a)(b-x)}$, it should give a semicircle which centre is $(\frac{a+b}{2},0)$

Evaluating Integrals with Modulus

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

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.

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.