June Revision Exercise 3 Q6

(a)(i)

Consider $y=\sqrt{a^2 - x^2} \Rightarrow y^2+x^2=a^2$ is a circle with centre Origin and radius $a$ .

Thus, required area is the area of a quadrant with radius $a$

$\text{Area} = \frac{1}{4} \pi a^2$

(ii)

$\int_0^a \frac{x^2}{\sqrt{a^2-x^2}} ~dx$

$= \int_0^2 x\frac{x}{\sqrt{a^2-x^2}} ~dx$

$= - x \sqrt{a^2-x^2} \Big|_0^a - \int_0^a -\sqrt{a^2-x^2} ~dx$

$= 0 + \frac{1}{4} \pi a^2$

$= \frac{1}{4} \pi a^2$

(b)

$u=3x+5 \Rightarrow \frac{du}{dx}=3$

$= \int_{-2}^{-\frac{5}{3}} (x-2)(3x+5)^3 ~dx$

$=\int_{-1}^0 (\frac{u-5}{3}-2)u^3 \frac{1}{4} ~du$

$=\frac{1}{9} \int_{-1}^0 u^4 - 11u^3 ~du$

$=\frac{1}{9} \Big| \frac{1}{5}u^5 - \frac{11}{4}u^4 \Big|_{-1}^0$

$= \frac{59}{180}$

Back to June Revision Exercise 3

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

June Revision Exercise 3 Q6

Leave a Comment Cancel reply

Leave a Comment
Cancel reply