2008 H2 Mathematics Paper 1 Question 1

Definite Integrals: Areas and Volumes

Answers

a=8143≈2.73{a = \sqrt[3]{\frac{81}{4}} \approx 2.73}

Full solutions

x=yx=\sqrt{y}
∫a4y  dy=∫01x2  dx[23y32]a4=[13x3]1223(432−a32)=13(23−13)23(8−a32)=738−a32=72a32=92a=8143  ■\begin{align*} \int_a^4 \sqrt{y} \; \mathrm{d}y & = \int_0^1 x^2 \; \mathrm{d}x \\ \left[ \frac{2}{3} y^{\frac{3}{2}} \right]_a^4 &= \left[ \frac{1}{3} x^3 \right]_1^2 \\ \frac{2}{3} \left( 4^{\frac{3}{2}} - a^{\frac{3}{2}} \right) &= \frac{1}{3} \left( 2^3 - 1^3 \right) \\ \frac{2}{3} \left( 8 - a^{\frac{3}{2}} \right) &= \frac{7}{3} \\ 8 - a^{\frac{3}{2}} &= \frac{7}{2} \\ a^{\frac{3}{2}} &= \frac{9}{2} \\ a &= \sqrt[3]{\frac{81}{4}} \; \blacksquare \end{align*}