June Revision Exercise 3 Q6

By KS Teng

(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

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