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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Integration techniques & volumes of revolution แพ็กฝึกเฉพาะหัวข้อ · เทคนิคปริพันธ์ · ปริมาตรทรงหมุน

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Substitution, parts, partial fractions & volumes, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Find $\displaystyle\int \frac{2x}{x^2+3}\,dx$.

Mark schemeshow ▾
M1 recognise the form $\displaystyle\int\frac{f'(x)}{f(x)}\,dx$ with $f(x)=x^2+3$  A1 $\ln(x^2+3)$  B1 $+\,c$. (Since $x^2+3>0$, no modulus is needed.)
Q2. 4 marks

Using the substitution $u=x^2-1$, find $\displaystyle\int 6x\,(x^2-1)^5\,dx$.

Mark schemeshow ▾
M1 $\dfrac{du}{dx}=2x$, so $du=2x\,dx$  M1 integral becomes $\displaystyle\int 3u^5\,du$  A1 $=\dfrac{u^6}{2}$  A1 $=\dfrac{(x^2-1)^6}{2}+c$.
Q3. 4 marks

Use integration by parts to find $\displaystyle\int x\,e^{2x}\,dx$.

Mark schemeshow ▾
M1 parts with $u=x$, $\dfrac{dv}{dx}=e^{2x}$  A1 $\dfrac{x}{2}e^{2x}-\displaystyle\int\dfrac12 e^{2x}\,dx$  A1 $=\dfrac{x}{2}e^{2x}-\dfrac14 e^{2x}$  B1 $+\,c$.
Q4. 5 marks

Express $\dfrac{x+7}{(x-1)(x+3)}$ in partial fractions, and hence find $\displaystyle\int \frac{x+7}{(x-1)(x+3)}\,dx$.

Mark schemeshow ▾
M1 $\dfrac{x+7}{(x-1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+3}$, so $x+7=A(x+3)+B(x-1)$  A1 $x=1$: $8=4A$, $A=2$  A1 $x=-3$: $4=-4B$, $B=-1$  M1 integrate each term to a log  A1 $2\ln|x-1|-\ln|x+3|+c$.
Q5. 5 marks

Use a suitable substitution to show that $\displaystyle\int_0^3 \frac{x}{1+x}\,dx = 3-2\ln 2$.

Mark schemeshow ▾
M1 let $u=1+x$, so $du=dx$ and $x=u-1$  A1 integrand becomes $\dfrac{u-1}{u}=1-\dfrac1u$  B1 limits: $x=0\to u=1$, $x=3\to u=4$  M1 $\displaystyle\Big[u-\ln u\Big]_1^4=(4-\ln 4)-(1-0)$  A1 $=3-\ln 4=3-2\ln 2$ ∎.
Q6. 6 marks

The region $R$ is bounded by the curve $y=e^{x}$, the $x$-axis and the lines $x=0$ and $x=1$. The region $R$ is rotated through $360°$ about the $x$-axis. Find the exact volume of the solid generated.

Mark schemeshow ▾
B1 $V=\pi\displaystyle\int_0^1 y^2\,dx$  B1 $y^2=e^{2x}$  M1 integrate: $\displaystyle\int e^{2x}\,dx$  A1 $=\dfrac12 e^{2x}$  M1 apply limits: $\pi\left(\dfrac{e^2}{2}-\dfrac12\right)$  A1 $V=\dfrac{\pi(e^2-1)}{2}$.
Q7. 6 marks

Use integration by parts to show that $\displaystyle\int_1^e x\ln x\,dx = \frac{e^2+1}{4}$.

Mark schemeshow ▾
B1 choose $u=\ln x$, $\dfrac{dv}{dx}=x$  M1 apply parts  A1 $\dfrac{x^2}{2}\ln x-\displaystyle\int\dfrac{x}{2}\,dx$  A1 $=\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}$  M1 apply limits: $\left(\dfrac{e^2}{2}-\dfrac{e^2}{4}\right)-\left(0-\dfrac14\right)$  A1 $=\dfrac{e^2}{4}+\dfrac14=\dfrac{e^2+1}{4}$ ∎.
Q8. 7 marks

(a) Express $\dfrac{3x+2}{x(x+2)}$ in partial fractions. (b) Hence show that $\displaystyle\int_1^2 \frac{3x+2}{x(x+2)}\,dx = \ln\frac{32}{9}$.

Mark schemeshow ▾
(a) M1 $\dfrac{3x+2}{x(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+2}$, so $3x+2=A(x+2)+Bx$  A1 $x=0$: $2=2A$, $A=1$  A1 $x=-2$: $-4=-2B$, $B=2$.
(b) M1 integrate each term to a log  A1 $\Big[\ln x+2\ln(x+2)\Big]_1^2$  M1 $=(\ln 2+2\ln 4)-(\ln 1+2\ln 3)$  A1 $=\ln\dfrac{2\times 16}{9}=\ln\dfrac{32}{9}$ ∎.
Q9. 7 marks

Using integration by parts twice, show that $\displaystyle\int_0^1 x^2 e^{x}\,dx = e-2$.

Mark schemeshow ▾
M1 first application, $u=x^2$: $x^2e^x-\displaystyle\int 2x\,e^x\,dx$  A1 correct  M1 second application, $u=2x$: $\displaystyle\int 2x\,e^x\,dx=2xe^x-\int 2e^x\,dx$  A1 $=2xe^x-2e^x$  A1 full antiderivative $e^x(x^2-2x+2)$  M1 apply limits: $e(1-2+2)-e^0(0-0+2)$  A1 $=e-2$ ∎.
Q10. 8 marks

The curve $C$ has equation $y=\dfrac{2}{\sqrt{2x+1}}$, $x\ge 0$. The region $R$ is bounded by $C$, the $x$-axis and the lines $x=0$ and $x=4$. (a) Show that the area of $R$ is $4$. (b) The region $R$ is rotated through $360°$ about the $x$-axis. Show that the volume of the solid generated is $4\pi\ln 3$.

Mark schemeshow ▾
(a) M1 write $y=2(2x+1)^{-\frac12}$ and integrate  A1 $\displaystyle\int 2(2x+1)^{-\frac12}\,dx=2(2x+1)^{\frac12}$  A1 $\Big[2\sqrt{2x+1}\Big]_0^4=6-2=4$ ∎.
(b) M1 $V=\pi\displaystyle\int_0^4 y^2\,dx$  A1 $y^2=\dfrac{4}{2x+1}$  M1 integrate: $\displaystyle\int\dfrac{4}{2x+1}\,dx=2\ln(2x+1)$  A1 $\pi\Big[2\ln(2x+1)\Big]_0^4=2\pi\ln 9$  A1 $=4\pi\ln 3$ ∎.