题目

Problem

(a) Using the substitution $u=\sqrt{2x+1}$ , show that

\int_4^{12}\sqrt{8x+4}\,e^{\sqrt{2x+1}}\,\mathrm{d}x

may be expressed in the form

\int_a^b ku^2e^u\,\mathrm{d}u

where $a$ , $b$ and $k$ are constants to be found.

(4)

(b) Hence find, by algebraic integration, the exact value of

\int_4^{12}\sqrt{8x+4}\,e^{\sqrt{2x+1}}\,\mathrm{d}x

giving your answer in simplest form.

(5)

题目中文翻译

(a) 令 $u=\sqrt{2x+1}$ ，证明

\int_4^{12}\sqrt{8x+4}\,e^{\sqrt{2x+1}}\,\mathrm{d}x

可写成

\int_a^b ku^2e^u\,\mathrm{d}u

的形式，其中 $a,b,k$ 为待求常数。

(4)

(b) 进而用代数积分法求

\int_4^{12}\sqrt{8x+4}\,e^{\sqrt{2x+1}}\,\mathrm{d}x

的精确值，并将答案化为最简形式。

(5)

解答

(a)

解法一

思路

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由指定代换 $u=\sqrt{2x+1}$ 求出微分关系，并把根式、指数及上下限同时换成 $u$ 。不能只把 $\mathrm{d}x$ 改写成 $\mathrm{d}u$ 而省略两者的关系。

答题过程

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Let $u=\sqrt{2x+1}$ . Then

u^2=2x+1.

Differentiating gives

2u\,\mathrm{d}u=2\,\mathrm{d}x,

\mathrm{d}x=u\,\mathrm{d}u.

Also,

\begin{align*} &\,\sqrt{8x+4}\\[4mm] =&\,\sqrt{4(2x+1)}\\[4mm] =&\,2u, \end{align*}

and $e^{\sqrt{2x+1}}=e^u$ .

When $x=4$ , $u=3$ ; when $x=12$ , $u=5$ . Thus, writing the given integral as $I$ ,

\begin{align*} I=&\,\int_3^5(2u)e^u\,u\,\mathrm{d}u\\[4mm] =&\,\int_3^5 2u^2e^u\,\mathrm{d}u. \end{align*}

Therefore,

\boxed{a=3,\quad b=5,\quad k=2}.

(b)

解法一

思路

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承接 (a)，对 $2u^2e^u$ 连续使用两次分部积分。先把多项式次数降低，再得到完整原函数；端点分别计算后相减，避免把两组长代值挤在同一行。

答题过程

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From part (a),

I=\int_3^5 2u^2e^u\,\mathrm{d}u.

For the first integration by parts, differentiate $2u^2$ and integrate $e^u$ . Writing

J=\int4ue^u\,\mathrm{d}u,

we obtain

\begin{align*} &\,\int2u^2e^u\,\mathrm{d}u\\[4mm] =&\,2u^2e^u-J. \end{align*}

Apply integration by parts again, differentiating $4u$ and integrating $e^u$ :

\begin{align*} J=&\,4ue^u-4\int e^u\,\mathrm{d}u\\[4mm] =&\,4ue^u-4e^u. \end{align*}

Hence an antiderivative is

\begin{align*} F(u)=&\,2u^2e^u-4ue^u\\[4mm] &\,\hspace{2pt}+4e^u\\[4mm] =&\,2e^u(u^2-2u+2). \end{align*}

At the endpoints,

\begin{align*} F(5)=&\,2e^5(25-10+2)\\ =&\,34e^5, \end{align*}

\begin{align*} F(3)=&\,2e^3(9-6+2)\\ =&\,10e^3. \end{align*}

Therefore,

\begin{align*} I=&\,F(5)-F(3)\\ =&\,\boxed{34e^5-10e^3}. \end{align*}

解法二

思路

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也可以用 D/I 表呈现分部积分。对 $2u^2$ 连续求导，对 $e^u$ 连续积分，再按正、负、正号组合对角乘积。多项式求导到零时停止；这等价于两次分部积分。

答题过程

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From part (a),

I=\int_3^5 2u^2e^u\,\mathrm{d}u.

Construct the D/I table:

Sign	D	I
$+$	$2u^2$	$e^u$
$-$	$4u$	$e^u$
$+$	$4$	$e^u$
-	$0$	$e^u$

Multiply each D entry by the I entry in the next row, using the indicated signs. The resulting antiderivative is

\begin{align*} F(u)=&\,2u^2e^u-4ue^u\\[4mm] &\,\hspace{2pt}+4e^u\\[4mm] =&\,2e^u(u^2-2u+2). \end{align*}

Thus

\begin{align*} F(5)=&\,2e^5(25-10+2)\\ =&\,34e^5, \end{align*}

\begin{align*} F(3)=&\,2e^3(9-6+2)\\ =&\,10e^3. \end{align*}

Therefore,

\begin{align*} I=&\,F(5)-F(3)\\ =&\,\boxed{34e^5-10e^3}. \end{align*}

P4 2023 Jan Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程