2019 H2 Maths Paper 2 Q1 — Worked Solution

Part (i)

2 marks

Since the integrand is a product of $x$ and $(1 - x)^{1/2}$ , we integrate by parts, taking $u = x$ (which differentiates to something simpler) and $dv = (1 - x)^{1/2}\, dx$ .

u = x,\quad\quad\frac{du}{dx} = 1

dv = (1 - x)^{1/2}\, dx,\quad\quad v = - \frac{2}{3}(1 - x)^{3/2}

By parts,

I

= uv - \int v\,\frac{du}{dx}\, dx

= - \frac{2}{3}x(1 - x)^{3/2} + \frac{2}{3}\int(1 - x)^{3/2}\, dx

= - \frac{2}{3}x(1 - x)^{3/2} - \frac{4}{15}(1 - x)^{5/2} + C

$\therefore I = - \frac{2}{3}x(1 - x)^{3/2} - \frac{4}{15}(1 - x)^{5/2} + C$ .

Part (ii)

4 marks

Using $u^{2} = 1 - x$ : differentiating, $2u\, du = - dx$ , so $dx = - 2u\, du$ ; also $x = 1 - u^{2}$ and $(1 - x)^{1/2} = u$ .

I = \int x(1 - x)^{1/2}\, dx

= \int\left( 1 - u^{2} \right)\, u\,( - 2u)\, du

= \int\left( 2u^{4} - 2u^{2} \right)du

= \frac{2}{5}u^{5} - \frac{2}{3}u^{3} + C

Since $u = (1 - x)^{1/2}$ ,

I = \frac{2}{5}(1 - x)^{5/2} - \frac{2}{3}(1 - x)^{3/2} + C

Part (iii)

Approach

Any two antiderivatives of the same function must differ by a constant; to see what that constant is here, we subtract the two expressions found in (i) and (ii) (without their constants of integration, which is where the difference would otherwise hide) and simplify by taking out the common factor $(1 - x)^{3/2}$ .

Let $I_{1} = - \frac{2}{3}x(1 - x)^{3/2} - \frac{4}{15}(1 - x)^{5/2}$ and $I_{2} = \frac{2}{5}(1 - x)^{5/2} - \frac{2}{3}(1 - x)^{3/2}$ , the two expressions with the constant dropped.

I_{1} - I_{2}

= - \frac{2}{3}x(1 - x)^{3/2} + \frac{2}{3}(1 - x)^{3/2} - \frac{4}{15}(1 - x)^{5/2} - \frac{2}{5}(1 - x)^{5/2}

= \frac{2}{3}(1 - x)^{3/2}\lbrack 1 - x\rbrack - \frac{2}{3}(1 - x)^{5/2}

= \frac{2}{3}(1 - x)^{5/2} - \frac{2}{3}(1 - x)^{5/2}

= 0

To check your answer: in factored form both (i) and (ii) reduce to

- \frac{2}{15}(3x + 2)(1 - x)^{3/2}

; differentiating this back gives

x(1 - x)^{1/2}

, the original integrand. ✓

$\therefore I_{1} - I_{2} = 0$ , a constant. (shown)

Note: the constant here happens to be $0$ — the two methods give exactly the same antiderivative, not merely antiderivatives that differ by a nonzero constant.

2019 H2 Maths Paper 2, Question 1

Part (i)

Part (ii)

Part (iii)

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