题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(i) (a) Divide $(x^2-3x+6)$ by $(x+2)$ .

(2)

(b) Hence find

\int \frac{x^2-3x+6}{x+2}\,dx

(3)

(ii) Using algebraic integration, find the exact value of

\int_0^{\pi/6}(\cos\theta-\sin\theta)^2\,d\theta

(5)

题目中文翻译

本题必须写出全部解题步骤。

不接受依赖计算器技术的解法。

(i) (a) 用 $(x+2)$ 去除 $(x^2-3x+6)$ 。

(b) 由此求

\int \frac{x^2-3x+6}{x+2}\,dx

(ii) 利用代数积分法求

\int_0^{\pi/6}(\cos\theta-\sin\theta)^2\,d\theta

的精确值。

解答

Use polynomial division:

\begin{array}{r|r} x+2 & x^2-3x+6 \end{array}

First,

x^2\div x=x

Then

x(x+2)=x^2+2x

Subtract:

x^2-3x+6-(x^2+2x)=-5x+6

Next,

-5x\div x=-5

Then

-5(x+2)=-5x-10

Subtract:

-5x+6-(-5x-10)=16

\boxed{\frac{x^2-3x+6}{x+2}=x-5+\frac{16}{x+2}}

The quotient is $x-5$ and the remainder is $16$ .

Using part (i)(a),

\int \frac{x^2-3x+6}{x+2}\,\mathrm{d}x =\int\left(x-5+\frac{16}{x+2}\right)\,\mathrm{d}x

Therefore

\begin{aligned} \int \frac{x^2-3x+6}{x+2}\,\mathrm{d}x &=\frac12x^2-5x+16\ln|x+2|+c \end{aligned}

\boxed{\int \frac{x^2-3x+6}{x+2}\,\mathrm{d}x =\frac12x^2-5x+16\ln|x+2|+c}

Expand the square first:

\begin{aligned} (\cos\theta-\sin\theta)^2 &=\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta \\ &=1-2\sin\theta\cos\theta \\ &=1-\sin 2\theta \end{aligned}

Therefore

\int_0^{\pi/6}(\cos\theta-\sin\theta)^2\,\mathrm{d}\theta =\int_0^{\pi/6}(1-\sin2\theta)\,\mathrm{d}\theta

Integrate:

\int(1-\sin2\theta)\,\mathrm{d}\theta =\theta+\frac12\cos2\theta

\begin{aligned} \int_0^{\pi/6}(\cos\theta-\sin\theta)^2\,\mathrm{d}\theta &=\left[\theta+\frac12\cos2\theta\right]_0^{\pi/6} \\ &=\left(\frac{\pi}{6}+\frac12\cos\frac{\pi}{3}\right) -\left(0+\frac12\cos0\right) \\ &=\left(\frac{\pi}{6}+\frac12\cdot\frac12\right) -\frac12 \\ &=\frac{\pi}{6}+\frac14-\frac12 \\ &=\frac{\pi}{6}-\frac14 \end{aligned}

Hence the exact value is

\boxed{\frac{\pi}{6}-\frac14}