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IAL 2021 Oct Q5

A Level / Edexcel / P3

IAL 2021 Oct Paper · Question 5

题目

Problem

(i) Find, by algebraic integration, the exact value of

∫248(2x−3)3 dx\int_2^4 \frac{8}{(2x-3)^3}\,dx
(4)

(ii) Find, in simplest form,

∫x(x2+3)7 dx\int x(x^2+3)^7\,dx
(2)
题目中文翻译

(i) 用代数积分求

∫248(2x−3)3 dx\int_2^4 \frac{8}{(2x-3)^3}\,dx

的精确值。

(ii) 求

∫x(x2+3)7 dx\int x(x^2+3)^7\,dx

的最简形式。

解答

(i)

We need to find

∫248(2x−3)3 dx\int_2^4 \frac{8}{(2x-3)^3}\,\mathrm{d}x

Write the integrand as

8(2x−3)−38(2x-3)^{-3}

Then

∫8(2x−3)−3 dx=8⋅(2x−3)−2−2⋅12\int 8(2x-3)^{-3}\,\mathrm{d}x =8\cdot\frac{(2x-3)^{-2}}{-2}\cdot\frac12

So

∫8(2x−3)−3 dx=−2(2x−3)−2\int 8(2x-3)^{-3}\,\mathrm{d}x =-2(2x-3)^{-2}

Therefore

∫248(2x−3)3 dx=[−2(2x−3)2]24=−252−(−212)=−225+2=4825\begin{aligned} \int_2^4 \frac{8}{(2x-3)^3}\,\mathrm{d}x &=\left[-\frac{2}{(2x-3)^2}\right]_2^4 \\ &=-\frac{2}{5^2}-\left(-\frac{2}{1^2}\right) \\ &=-\frac{2}{25}+2 \\ &=\frac{48}{25} \end{aligned}

Hence the exact value is

4825\boxed{\frac{48}{25}}

(ii)

Let

u=x2+3u=x^2+3

Then

dudx=2x\frac{\mathrm{d}u}{\mathrm{d}x}=2x

so

x dx=12 dux\,\mathrm{d}x=\frac12\,\mathrm{d}u

Therefore

∫x(x2+3)7 dx=12∫u7 du\int x(x^2+3)^7\,\mathrm{d}x =\frac12\int u^7\,\mathrm{d}u

So

12∫u7 du=12⋅u88+c=u816+c\frac12\int u^7\,\mathrm{d}u =\frac12\cdot\frac{u^8}{8}+c =\frac{u^8}{16}+c

Substituting back,

∫x(x2+3)7 dx=(x2+3)816+c\boxed{\int x(x^2+3)^7\,\mathrm{d}x=\frac{(x^2+3)^8}{16}+c}