题目

Problem

(i) Find

\begin{align*} \int\frac{(3x+2)^2}{4\sqrt{x}}\,dx, \qquad x>0 \end{align*}

giving your answer in simplest form.

(5)

(ii) A curve $C$ has equation $y=f(x)$ .

Given

find, in simplest form, $f(x)$ .

(6)

解答

展开

先展开分子，再除以 $4\sqrt{x}$ ，把被积函数拆成幂函数。

展开

\begin{align*} (3x+2)^2=9x^2+12x+4. \end{align*}

\begin{align*} \frac{(3x+2)^2}{4\sqrt{x}} =&\,\frac{9x^2+12x+4}{4x^{1/2}}\\ =&\,\frac94x^{3/2}+3x^{1/2}+x^{-1/2}. \end{align*}

Therefore

\begin{align*} \int\frac{(3x+2)^2}{4\sqrt{x}}\,dx =&\,\int\left(\frac94x^{3/2}+3x^{1/2}+x^{-1/2}\right)\,dx\\ =&\,\frac94\cdot\frac{x^{5/2}}{5/2} +3\cdot\frac{x^{3/2}}{3/2} +\frac{x^{1/2}}{1/2} +c\\ =&\,\frac9{10}x^{5/2}+2x^{3/2}+2x^{1/2}+c. \end{align*}

展开

先由 $f'(3)=2$ 得到一个关于 $a,b$ 的方程。再把 $f'(x)$ 积分成 $f(x)$ ，利用 $y$ 轴截距确定常数为 $-8$ ，再用点 $P(3,-2)$ 得到第二个方程。

展开

Since the gradient at $P(3,-2)$ is $2$ ,

\begin{align*} f'(3)=&\,2\\ 3^2+3a+b=&\,2\\ 3a+b=&\,-7. \end{align*}

Integrate $f'(x)$ :

\begin{align*} f(x) =&\,\int(x^2+ax+b)\,dx\\ =&\,\frac13x^3+\frac12ax^2+bx+c. \end{align*}

The $y$ -intercept is $-8$ , so

\begin{align*} c=-8. \end{align*}

Thus

\begin{align*} f(x)=\frac13x^3+\frac12ax^2+bx-8. \end{align*}

Use $P(3,-2)$ :

\begin{align*} -2 =&\,\frac13(3)^3+\frac12a(3)^2+3b-8\\ -2=&\,9+\frac92a+3b-8\\ \frac92a+3b=&\,-3. \end{align*}

So we solve

\begin{align*} 3a+b=&\,-7,\\ \frac92a+3b=&\,-3. \end{align*}

From the first equation,

\begin{align*} b=-7-3a. \end{align*}

Substitute:

\begin{align*} \frac92a+3(-7-3a)=&\,-3\\ \frac92a-21-9a=&\,-3\\ -\frac92a=&\,18\\ a=&\,-4. \end{align*}

Then

\begin{align*} b=-7-3(-4)=5. \end{align*}

Therefore

\begin{align*} f(x)=\frac13x^3-2x^2+5x-8. \end{align*}