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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Integration แพ็กฝึกเฉพาะหัวข้อ · พื้นที่และกฎสี่เหลี่ยมคางหมู

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Power rule, areas & trapezium rule, 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 $\displaystyle\int \left(6x^2-4x+3\right)\,dx$.

Mark schemeshow ▾
M1 raise each power by 1 and divide by the new power  A1 $2x^3-2x^2$  A1 $2x^3-2x^2+3x+c$ (all terms and $+c$).
Q2. 4 marks

Find $\displaystyle\int \left(6\sqrt{x}-\frac{4}{x^2}\right)dx$, giving each term in its simplest form.

Mark schemeshow ▾
M1 rewrite as $6x^{1/2}-4x^{-2}$  M1 apply the power rule to at least one term  A1 $6\cdot\tfrac{x^{3/2}}{3/2}=4x^{3/2}$  A1 $-4\cdot\tfrac{x^{-1}}{-1}=+\tfrac{4}{x}$, so $4x^{3/2}+\tfrac{4}{x}+c$.
Q3. 4 marks

Evaluate $\displaystyle\int_{1}^{4} \left(3\sqrt{x}+1\right)dx$.

Mark schemeshow ▾
M1 integrate: $3x^{1/2}\to 2x^{3/2}$  A1 $\left[2x^{3/2}+x\right]_1^4$  M1 substitute limits: $(2\cdot 8+4)-(2+1)$  A1 $20-3=17$.
Q4. 5 marks

A curve has gradient function $\dfrac{dy}{dx}=6x^2-8x+3$ and passes through the point $(2,\,7)$. Find the equation of the curve.

Mark schemeshow ▾
M1 integrate the gradient function  A1 $y=2x^3-4x^2+3x$  B1 $+c$ included  M1 substitute $(2,7)$: $16-16+6+c=7$  A1 $c=1$, so $y=2x^3-4x^2+3x+1$.
Q5. 5 marks

Evaluate $\displaystyle\int_{1}^{2} \left(4x^3-\frac{6}{x^3}\right)dx$, giving your answer as an exact fraction.

Mark schemeshow ▾
M1 rewrite as $4x^3-6x^{-3}$  M1 integrate: $-6x^{-3}\to+3x^{-2}$  A1 $\left[x^4+\tfrac{3}{x^2}\right]_1^2$  M1 substitute limits: $\left(16+\tfrac34\right)-(1+3)$  A1 $=\tfrac{67}{4}-4=\tfrac{51}{4}$.
Q6. 6 marks

The curve $y=6x-x^2$ crosses the $x$-axis at the origin $O$ and at the point $A$. (a) Find the coordinates of $A$. (b) Find the area of the region enclosed by the curve and the $x$-axis.

Mark schemeshow ▾
(a) M1 $x(6-x)=0$  A1 $A(6,\,0)$.
(b) M1 area $=\displaystyle\int_0^6 (6x-x^2)\,dx$  A1 $\left[3x^2-\tfrac{x^3}{3}\right]_0^6$  M1 substitute limits: $108-72$  A1 $=36$.
Q7. 6 marks

The table gives values of $y=\sqrt{1+x^3}$, correct to 3 decimal places, for $x=0,\ 0.5,\ 1,\ 1.5,\ 2$: the known values are $y=1$ at $x=0$, $y=1.414$ at $x=1$, and $y=3$ at $x=2$. (a) Find the missing values of $y$ at $x=0.5$ and $x=1.5$, to 3 decimal places. (b) Use the trapezium rule with all five values to estimate $\displaystyle\int_0^2 \sqrt{1+x^3}\,dx$. (c) The curve is convex (it bends upwards) on this interval. State, with a reason, whether your estimate is an overestimate or an underestimate.

Mark schemeshow ▾
(a) B1 $y(0.5)=\sqrt{1.125}=1.061$  B1 $y(1.5)=\sqrt{4.375}=2.092$.
(b) M1 $\tfrac{h}{2}\left[y_0+y_4+2(y_1+y_2+y_3)\right]$ with $h=0.5$  A1 $\tfrac{0.5}{2}\left[1+3+2(1.061+1.414+2.092)\right]$  A1 $\approx 3.283$.
(c) B1 overestimate — the chords of a convex curve lie above the curve, so each trapezium contains extra area.
Q8. 7 marks

The curve $y=x^2-4x+8$ and the line $y=x+4$ intersect at two points. Find the exact area of the finite region enclosed between the curve and the line.

Mark schemeshow ▾
M1 $x^2-4x+8=x+4 \Rightarrow x^2-5x+4=0$  A1 $x=1,\ 4$  M1 area $=\displaystyle\int_1^4\left[(x+4)-(x^2-4x+8)\right]dx$  A1 $=\displaystyle\int_1^4(5x-x^2-4)\,dx$  M1 integrate: $\left[\tfrac{5x^2}{2}-\tfrac{x^3}{3}-4x\right]_1^4$  A1 $=\tfrac{8}{3}-\left(-\tfrac{11}{6}\right)$  A1 $=\tfrac{9}{2}$.
Q9. 7 marks

$f(x)=x^3-4x$. (a) Write down the values of $x$ with $0\le x\le 3$ at which the curve $y=f(x)$ crosses the $x$-axis. (b) Show that $\displaystyle\int_0^3 f(x)\,dx=\tfrac{9}{4}$. (c) Find the total area of the region enclosed between the curve and the $x$-axis for $0\le x\le 3$, explaining why your answer is not $\tfrac{9}{4}$.

Mark schemeshow ▾
(a) B1 $x=0$ and $x=2$ (curve is below the axis between them).
(b) M1 $\left[\tfrac{x^4}{4}-2x^2\right]_0^3$  A1 $=\tfrac{81}{4}-18=\tfrac{9}{4}$ ∎.
(c) M1 split at $x=2$: area $=\left|\int_0^2 f\right|+\int_2^3 f$  A1 $\int_0^2 f\,dx=4-8=-4$, giving area $4$  A1 $\int_2^3 f\,dx=\tfrac{9}{4}-(-4)=\tfrac{25}{4}$  A1 total area $=4+\tfrac{25}{4}=\tfrac{41}{4}$; the single integral $\tfrac94$ is smaller because the region below the axis counts as negative and cancels part of the region above.
Q10. 8 marks

The region $R$ is bounded by the curve $y=\sqrt{x}$, the line $y=x-2$ and the $x$-axis. (a) Show that the curve and the line intersect at the point $(4,\,2)$. (b) Find the exact area of $R$.

Mark schemeshow ▾
(a) M1 $\sqrt{x}=x-2$; let $u=\sqrt{x}$: $u^2-u-2=0$  A1 $(u-2)(u+1)=0$, $u=2$ (reject $u=-1$ since $\sqrt{x}\ge 0$)  A1 $x=4$, $y=2$, i.e. $(4,2)$ ∎.
(b) M1 area under curve: $\displaystyle\int_0^4 x^{1/2}\,dx=\left[\tfrac{2}{3}x^{3/2}\right]_0^4$  A1 $=\tfrac{16}{3}$  B1 line meets the $x$-axis at $x=2$, so the triangle under the line has area $\tfrac12\cdot 2\cdot 2=2$  M1 area of $R$ = area under curve − area of triangle  A1 $=\tfrac{16}{3}-2=\tfrac{10}{3}$.