Every Integration: Areas and the Trapezium Rule question from the WMA12 papers in our bank, 2019–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
Level is estimated from marks: 1–6 easy, 7–8 medium, 9+ hard.
The $x$ values go from $4.0$ to $5.0$ in steps of $0.2$, so $h=0.2$.
$\displaystyle \int_4^5 f(x)\,dx\approx \dfrac{0.2}{2}\left[9.2+8.6+2(8.4556+3.8512+5.0342+7.8297)\right]$.
The value is $6.81414$, so the approximation is $6.814$ to 3 decimal places.
$6.814$.
$\int\left(\dfrac{12}{x^2}+4\right)\,dx=\int(12x^{-2}+4)\,dx=-\dfrac{12}{x}+4x$.
$\left[-\dfrac{12}{x}+4x\right]_k^{2k}=\left(-\dfrac6k+8k\right)-\left(-\dfrac{12}{k}+4k\right)=\dfrac6k+4k$.
Given the integral is $14$, $\dfrac6k+4k=14$. Multiplying by $k$ gives $4k^2-14k+6=0$, so $2k^2-7k+3=0$.
$(2k-1)(k-3)=0$, hence $k=\dfrac12$ or $k=3$. Both are positive, so both are valid.
$k=\dfrac12$ or $k=3$.
The gradient of $PQ$ is $\dfrac{5-11}{4-1}=-2$, so the line through $P(1,11)$ is $y-11=-2(x-1)$.
Thus $PQ$ has equation $y=-2x+13$.
From $x=1$ to $x=4$, the line segment is above the curve. Hence $R=\displaystyle\int_1^4\left[(-2x+13)-\left(3x+\dfrac{16}{x^2}-8\right)\right]dx$.
$R=\displaystyle\int_1^4(21-5x-16x^{-2})\,dx$.
$R=\left[21x-\dfrac52x^2+\dfrac{16}{x}\right]_1^4$.
$R=(84-40+4)-\left(21-\dfrac52+16\right)=48-\dfrac{69}{2}=\dfrac{27}{2}$.
$\dfrac{27}{2}$.
The $x$ values increase by $1.25$, so the trapezium width is $h=1.25$.
$\displaystyle \int_{0.5}^{5.5} f(x)\,dx\approx\dfrac{1.25}{2}\{3.479+5.182+2(6.101+7.448+6.823)\}$.
This gives $30.878125$, which is $30.9$ to one decimal place.
$\displaystyle \int_{0.5}^{5.5}(f(x)+4x)\,dx=\int_{0.5}^{5.5}f(x)\,dx+\int_{0.5}^{5.5}4x\,dx$.
$\displaystyle \int_{0.5}^{5.5}4x\,dx=[2x^2]_{0.5}^{5.5}=2(5.5)^2-2(0.5)^2=60$.
Using the estimate from part (a), the required value is $30.9+60=90.9$.
$(a)$ $30.9$. $\quad (b)$ $90.9$.
The $x$ values increase by $1.5$, so $h=1.5$.
$19.3=\dfrac{1.5}{2}[4.16+2.28+2(2.91+a+1.73+1.37+1.43)]$.
This simplifies to $19.3=15.99+1.5a$, so $a=\dfrac{3.31}{1.5}=2.206\ldots$.
Therefore $a=2.21$ to 3 significant figures.
$\displaystyle \int_{-4}^{5}(2f(x)-3)dx=2\int_{-4}^{5}f(x)dx-\int_{-4}^{5}3\,dx$.
This is $2(19.3)-3(9)=38.6-27=11.6$.
$(a)$ $a=2.21$ to 3 s.f. $\quad (b)$ $11.6$.
The $x$ values increase by $3$, so $h=3$.
$A\approx \dfrac32\{1.3195+0.4665+2(1.0718+0.8706+0.7071+0.5743)\}$.
$A\approx \dfrac32(8.2336)=12.3504$, so the area is $12.35$ to 2 decimal places.
A more accurate estimate would use more trapezia, so the strip width is smaller.
Since $2^{3-0.1x}=8\cdot2^{-0.1x}$, the required estimate is $8(12.35)=98.8$.
$(a)\ 12.35.\quad (b)$ Use narrower strips, or more trapezia. $\quad (c)\ 98.8$.
The $x$ values increase by $1.5$, so $h=1.5$.
$\displaystyle\int_{-2}^{7}3\left(\dfrac12\right)^x dx\approx\dfrac{1.5}{2}\{12+0.023+2(4.243+1.5+0.530+0.188+0.066)\}$.
This gives $18.80775$, which is $18.8$ to one decimal place.
$3\left(\dfrac12\right)^{x+2}=\dfrac14\cdot3\left(\dfrac12\right)^x$.
So $\displaystyle\int_{-2}^{7}3\left(\dfrac12\right)^{x+2}dx\approx\dfrac14(18.80775)=4.7019375$, which is $4.7$.
Since $3\left(\dfrac12\right)^x=3\cdot2^{-x}$, we have $2^{-x}=\dfrac13\cdot3\left(\dfrac12\right)^x$.
$\displaystyle\int_{-2}^{7}(2^{-x}+2x)dx\approx\dfrac{18.80775}{3}+[x^2]_{-2}^{7}=6.26925+45=51.26925$, which is $51.3$.
$(a)$ $18.8$. $\quad (b)(i)$ $4.7$. $\quad (b)(ii)$ $51.3$.
The $x$ values are $0,3,6,9,12,15$, so the trapezium width is $h=3$.
$\text{Area}\approx \dfrac32\left[0+0+2(1.52+2.74+3.12+3.08)\right]$.
This gives $31.38\text{ m}^2$.
The speed is $1.5\text{ m s}^{-1}$, so in one minute the water travels $1.5\times60=90\text{ m}$.
Volume per minute $\approx31.38\times90=2824.2\text{ m}^3$.
To 2 significant figures, this is $2.8\times10^3\text{ m}^3$.
It is an underestimate, because the curve between the plotted points lies below the straight-line trapezium chords in the diagram, so the true cross-sectional area is larger.
$(a)$ $31.38\text{ m}^2$. $\quad (b)$ $2.8\times10^3\text{ m}^3$. $\quad (c)$ Underestimate.
The $x$ values run from $-1$ to $1$ in steps of $0.5$, so $h=0.5$.
$\displaystyle \int_{-1}^{1} f(x)\,dx\approx \dfrac{0.5}{2}\left[2.287+2.834+2(4.470+6.719+7.291)\right]$.
This gives $10.52025$, so the estimate is $10.52$ to 2 decimal places.
$\displaystyle \int_{-1}^{1}(f(x)-2)\,dx=\int_{-1}^{1}f(x)\,dx-\int_{-1}^{1}2\,dx$.
The interval has width $2$, so $\int_{-1}^{1}2\,dx=4$. Hence the estimate is $10.52-4=6.52$.
For $\displaystyle \int_{1}^{3} f(x-2)\,dx$, let $u=x-2$. When $x=1$, $u=-1$, and when $x=3$, $u=1$.
So $\displaystyle \int_{1}^{3} f(x-2)\,dx=\int_{-1}^{1} f(u)\,du$, which has the same estimate $10.52$.
$(a)$ $10.52$. $\quad (b)(i)\ 6.52. $\quad $(b)(ii)\ 10.52$.
$P(10)=0.95+2^{-2}+2^2-(10-12)^2=0.95+0.25+4-4=1.20$.
$P(11.5)=0.95+2^{-0.5}+2^{0.5}-(11.5-12)^2=2.821\ldots$.
The time values increase by $0.5$, so $h=0.5$.
The estimate is $\dfrac{0.5}{2}\{1.20+2.95+2(1.882+2.45+2.821\ldots)\}$.
This gives $4.614\ldots$, so the estimated energy is $4.61\text{ kWh}$ to 2 decimal places.
$(a)\ P(10)=1.20,\ P(11.5)=2.821.\quad (b)\ 4.61\text{ kWh}$.
The curve is $y=2x^2+7$. Set $2x^2+7=17$ to get $x^2=5$, so the right-hand limit is $x=\sqrt5$.
The shaded area is between $y=17$ and $y=2x^2+7$, from $x=0$ to $x=\sqrt5$.
$A=\int_0^{\sqrt5}\{17-(2x^2+7)\}\,dx=\int_0^{\sqrt5}(10-2x^2)\,dx$.
$A=\left[10x-\dfrac23x^3\right]_0^{\sqrt5}=10\sqrt5-\dfrac23(5\sqrt5)=\dfrac{20\sqrt5}{3}$.
$\dfrac{20\sqrt5}{3}$.
Since $P(k,300000)$ lies on $y=3\times2^{-x}$, $300000=3\times2^{-k}$.
$100000=2^{-k}$, so $-k=\log_2(100000)$ and $k=-\log_2(100000)=-16.609\ldots$.
To $2$ decimal places, $k=-16.61$.
The $x$ values increase by $1.5$, so $h=1.5$.
$\displaystyle\int_{-0.5}^{7}3\times2^{-x}dx\approx\dfrac{1.5}{2}\{4.243+0.023+2(1.5+0.530+0.188+0.066)\}$.
This gives $6.6255$, which is $6.63$ to $2$ decimal places.
$\displaystyle\int_{-0.5}^{7}2^{-x}dx=\dfrac13\int_{-0.5}^{7}3\times2^{-x}dx$.
Using the substitution $u=-x$, $\displaystyle\int_{-7}^{0.5}2^x dx=\int_{-0.5}^{7}2^{-u}du$.
The required estimate is therefore $\dfrac23(6.6255)=4.417$, which is $4.42$ to $2$ decimal places.
$(a)$ $k=-16.61$. $\quad (b)(i)$ $6.63$. $\quad (b)(ii)$ $4.42$.
In $(3+px)^5$, the constant term is $3^5=243$, so $A=243$.
The coefficient of $x$ is $5(3^4)p=405p$, so $B=405p$.
The coefficient of $x^3$ is $\binom53(3^2)p^3=10\cdot9p^3=90p^3$, so $D=90p^3$.
$405p=18(90p^3)=1620p^3$.
Since $p<0$, $p\ne0$. Divide by $405p$ to get $1=4p^2$, so $p=\pm\dfrac12$. Therefore $p=-\dfrac12$.
The coefficient of $x^2$ is $\binom52(3^3)p^2=10\cdot27\cdot\dfrac14=\dfrac{135}{2}$.
$(a)$ $A=243$. $\quad (b)(i)$ $p=-\dfrac12$. $\quad (b)(ii)$ $C=\dfrac{135}{2}$.
The $x$ values increase by $0.5$, so the trapezium-rule width is $h=0.5$.
$\displaystyle \int_0^2 f(x)\,dx\approx \dfrac{0.5}{2}\left[3+1.92+2(2.6833+2.4+2.1466)\right]$.
This gives $4.84495$, so the approximation is $4.845$ to 3 decimal places.
$\displaystyle \int_0^2 \left(2-\dfrac14x^2\right)dx=\left[2x-\dfrac{x^3}{12}\right]_0^2=4-\dfrac{2}{3}=\dfrac{10}{3}$.
The shaded region lies between $C_1$ and $C_2$, so its estimated area is $4.845-\dfrac{10}{3}=1.5116\ldots$.
The rectangle has area $2\times3=6$.
Percentage shaded $=\dfrac{1.5116\ldots}{6}\times100=25.19\ldots\%$, so the estimate is $25.2\%$.
$(a)$ $4.845$. $\quad (b)\ 25.2\%$.
Since $(x+2)$ is a factor, $f(-2)=0$.
$A(-2)^3+6(-2)^2-4(-2)+B=0$, so $-8A+32+B=0$ and $B=8A-32$.
$\int_3^5(Ax^3+6x^2-4x+B)\,dx=176$.
$\left[\dfrac A4x^4+2x^3-2x^2+Bx\right]_3^5=176$.
This gives $136A+164+2B=176$, so $68A+B=6$.
Using $B=8A-32$ in $68A+B=6$ gives $76A=38$, so $A=\dfrac12$.
$B=8\left(\dfrac12\right)-32=-28$.
$A=\dfrac12,\quad B=-28$.
The $x$ values increase by $0.5$, so $h=0.5$.
$\int_5^7 2^{5-\sqrt{x}}\,dx\approx\dfrac{0.5}{2}\{6.792+5.113+2(6.298+5.858+5.466)\}$.
This gives $11.78725$, so the estimate is $11.79$ to 2 decimal places.
$2^{6-\sqrt{x}}=2\cdot2^{5-\sqrt{x}}$, so $\int_5^7 2^{6-\sqrt{x}}\,dx\approx2(11.78725)=23.5745$.
This is $23.57$ to 2 decimal places.
$\int_5^7(3+2^{5-\sqrt{x}})\,dx=\int_5^7 3\,dx+\int_5^7 2^{5-\sqrt{x}}\,dx$.
The estimate is $3(7-5)+11.78725=17.78725$, so $17.79$ to 2 decimal places.
$(a)\ 11.79.\quad (b)(i)\ 23.57.\quad (b)(ii)\ 17.79$.
$x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$.
The circle becomes $(x-3)^2+(y+4)^2=25-k$.
The centre is $(3,-4)$. Since the $x$-axis is tangent, the radius is the distance from the centre to the $x$-axis, which is $4$.
Thus $25-k=4^2=16$, so $k=9$.
With $k=9$, the circle is $(x-3)^2+(y+4)^2=16$. On the $x$-axis, $y=0$, giving $x=3$.
On the $y$-axis, $x=0$, so $9+(y+4)^2=16$. Hence $y=-4\pm\sqrt7$.
The vertical base on the $y$-axis has length $2\sqrt7$, and its perpendicular distance to $(3,0)$ is $3$. The area is $\dfrac12(2\sqrt7)(3)=3\sqrt7$.
$(a)\ k=9.\quad (b)\ 3\sqrt7$.
The strip width is $h=3$. Therefore $\int_2^{14}\log_2(2x)\,dx\approx\dfrac32\{2+4.81+2(3.32+4+4.46)\}$.
The estimate is $\dfrac32(30.37)=45.555$, so to one decimal place the answer is $45.6$.
$\log_2(4x^2)=\log_2\{(2x)^2\}=2\log_2(2x)$, so $\int_2^{14}\dfrac{\log_2(4x^2)}5\,dx=\dfrac25\int_2^{14}\log_2(2x)\,dx$.
Using part (a), the estimate is $\dfrac25(45.555)=18.222$, so $18.2$ to one decimal place.
$\log_2\left(\dfrac2x\right)=2-\log_2(2x)$.
$\int_2^{14}\log_2\left(\dfrac2x\right)dx\approx\int_2^{14}2\,dx-45.555=24-45.555=-21.555$, so $-21.6$.
$(a)\ 45.6.\quad (b)(i)\ 18.2.\quad (b)(ii)\ -21.6$.
For $0<a<1$, $y=a^x$ is a decreasing exponential curve. At $x=0$, $y=a^0=1$, so it crosses the $y$-axis at $(0,1)$.
The $x$ values increase by $0.5$, so $h=0.5$.
$A\approx\dfrac{0.5}{2}\{4.25+16.06+2(6.427+9.125+12.34)\}$.
$A\approx0.25(76.094)=19.0235$, so the estimate is $19.0$ to 3 significant figures.
$x(x-3)+\left(\dfrac12\right)^x=x^2+\left(\dfrac12\right)^x-3x$.
$\int_2^4 3x\,dx=\left[\dfrac32x^2\right]_2^4=24-6=18$.
Using the estimate from part (b), the required value is $19.0235-18=1.0235$, so the estimate is $1.0$.
$(a)$ A decreasing exponential curve crossing the $y$-axis at $(0,1)$. $\quad (b)\ 19.0.\quad (c)\ 1.0$.
$y=\dfrac{\sqrt{x}(100-x^2)}{40}=\dfrac52x^{1/2}-\dfrac1{40}x^{5/2}$.
$\dfrac{dy}{dx}=\dfrac54x^{-1/2}-\dfrac1{16}x^{3/2}$.
At $P$, $\dfrac{dy}{dx}=0$. Thus $\dfrac54x^{-1/2}=\dfrac1{16}x^{3/2}$. Multiplying by $16x^{1/2}$ gives $20=x^2$.
Since $x\geq0$, $x=\sqrt{20}=2\sqrt5$.
$\int y\,dx=\int\left(\dfrac52x^{1/2}-\dfrac1{40}x^{5/2}\right)dx=\dfrac53x^{3/2}-\dfrac1{140}x^{7/2}$.
The signed area from $0$ to $k$ is zero because the positive area $R_1$ equals the negative area $R_2$ in magnitude.
$\dfrac53k^{3/2}-\dfrac1{140}k^{7/2}=0$. Since $k>0$, divide by $k^{3/2}$ to get $\dfrac53-\dfrac{k^2}{140}=0$, so $k^2=\dfrac{700}{3}$.
$k=\sqrt{\dfrac{700}{3}}=\dfrac{10\sqrt{21}}{3}$.
$(a)\ x=2\sqrt5.\quad (b)\ k=\dfrac{10\sqrt{21}}{3}$.
The width is $1$, so $\displaystyle\int_{-1}^{4}\log_{10}(x+5)\,dx\approx\dfrac12\{\log_{10}4+\log_{10}9+2(\log_{10}5+\log_{10}6+\log_{10}7+\log_{10}8)\}$.
This is $\log_{10}\left(\sqrt{4\times9}\times5\times6\times7\times8\right)$.
$\sqrt{4\times9}\times5\times6\times7\times8=6\times1680=10080$, so $k=10080$.
The logarithms require $5-a>0$ and $a+25>0$, so $-25<a<5$.
$2\log_5(5-a)-\log_5(a+25)=\log_5\left(\dfrac{(5-a)^2}{a+25}\right)$.
$1=\log_5 5$, so $\dfrac{(5-a)^2}{a+25}=5$.
$(5-a)^2=5(a+25)$, so $a^2-10a+25=5a+125$. Hence $a^2-15a-100=0$.
$(a-20)(a+5)=0$, so $a=20$ or $a=-5$. Only $a=-5$ is in the domain, so $a=-5$.
$(i)$ $k=10080$. $\quad (ii)$ $a=-5$.
$x^2+3=13-\dfrac9{x^2}$.
Multiplying by $x^2$ gives $x^4+3x^2=13x^2-9$, so $x^4-10x^2+9=0$.
Let $u=x^2$. Then $u^2-10u+9=(u-1)(u-9)=0$, so $x^2=1$ or $x^2=9$.
Since $x>0$, the intersections have $x=1$ and $x=3$. Thus $x_P=1$ and $x_Q=3$.
Between $x=1$ and $x=3$, curve $C_2$ is above curve $C_1$, so $R=\displaystyle\int_1^3\left(13-\dfrac9{x^2}-(x^2+3)\right)dx$.
$R=\displaystyle\int_1^3(10-x^2-9x^{-2})\,dx$.
$R=\left[10x-\dfrac{x^3}{3}+\dfrac9x\right]_1^3$.
$R=\left(30-9+3\right)-\left(10-\dfrac13+9\right)=24-\dfrac{56}{3}=\dfrac{16}{3}$.
$(a)$ $x_P=1$, $x_Q=3$. $\quad (b)$ $\dfrac{16}{3}$.
$y=2x^{3/2}(4-x)=8x^{3/2}-2x^{5/2}$.
$\dfrac{dy}{dx}=12x^{1/2}-5x^{3/2}=\sqrt{x}(12-5x)$.
For the interior stationary point $P$, $12-5x=0$, so $x=\dfrac{12}{5}$.
$R_1$ is the area from $x=0$ to $x=4$ above the $x$-axis. $R_2$ is the area from $x=4$ to $x=k$ below the $x$-axis.
Equal areas mean $\displaystyle \int_0^4 y\,dx=-\int_4^k y\,dx$, so $\displaystyle \int_0^k y\,dx=0$.
$\displaystyle \int 2x^{3/2}(4-x)dx=\int(8x^{3/2}-2x^{5/2})dx=\dfrac{16}{5}x^{5/2}-\dfrac47x^{7/2}$.
$\dfrac{16}{5}k^{5/2}-\dfrac47k^{7/2}=0$.
$4k^{5/2}\left(\dfrac45-\dfrac{k}{7}\right)=0$.
Since $k>4$, $\dfrac45-\dfrac{k}{7}=0$, so $k=\dfrac{28}{5}$.
$(a)$ $x=\dfrac{12}{5}$. $\quad (b)$ $k=\dfrac{28}{5}$.
At the $y$-axis, $x=0$. On $C$, $y=0^2-4(0)+5=5$, so $D=(0,5)$.
The line is $y=2$, so solve $x^2-4x+5=2$.
$x^2-4x+3=0$, so $(x-1)(x-3)=0$. Hence $x_E=1$ and $x_F=3$.
$R_1$ lies between the curve and the line $y=2$ from $x=0$ to $x=1$.
$\displaystyle R_1=\int_0^1\left(x^2-4x+5-2\right)dx=\int_0^1(x^2-4x+3)dx$.
$\displaystyle R_1=\left[\dfrac{x^3}{3}-2x^2+3x\right]_0^1=\dfrac13-2+3=\dfrac43$.
The line through $D=(0,5)$ and $F=(3,2)$ has gradient $-1$, so its equation is $y=5-x$.
For $0\le x\le1$, $R_2$ is between $y=5-x$ and the curve. For $1\le x\le3$, it is between $y=5-x$ and $y=2$.
$\displaystyle R_2=\int_0^1\left[(5-x)-(x^2-4x+5)\right]dx+\int_1^3\left[(5-x)-2\right]dx$.
$\displaystyle R_2=\int_0^1(-x^2+3x)dx+\int_1^3(3-x)dx=\dfrac{7}{6}+2=\dfrac{19}{6}$.
$k=\dfrac{\text{area of }R_1}{\text{area of }R_2}=\dfrac{4/3}{19/6}=\dfrac{8}{19}$.
$(a)$ $D=(0,5)$. $\quad (b)$ $x_E=1,\ x_F=3$. $\quad (c)$ $k=\dfrac{8}{19}$.
Using $y=1-\log_{10}(\sin x)$ gives $y(1)=1.075$ and $y(2)=1.041$ to 3 decimal places.
The $x$ values increase by $0.5$, so $h=0.5$.
$A\approx\dfrac{0.5}{2}\{1.319+1.850+2(1.075+1.001+1.041+1.223)\}$.
This gives $2.9625\ldots$, so the estimate is $2.96$ to 2 decimal places.
Since $y=1-\log_{10}(\sin x)$, we have $\log_{10}(\sin x)=1-y$.
$3+\log_{10}(\sin x)=4-y$, and the interval from $0.5$ to $3$ has length $2.5$.
$\int_{0.5}^{3}(3+\log_{10}(\sin x))\,dx\approx4(2.5)-2.9625\ldots=7.0374\ldots$, so $7.04$.
$(a)\ y(1)=1.075,\ y(2)=1.041.\quad (b)\ 2.96.\quad (c)\ 7.04$.
Four equal strips from $x=2$ to $x=14$ have width $h=\dfrac{14-2}{4}=3$.
Use $x=2,5,8,11,14$ and $y=\log_{10}x$.
$A\approx\dfrac32\{\log_{10}2+\log_{10}14+2(\log_{10}5+\log_{10}8+\log_{10}11)\}=10.10\ldots$.
Use more strips of smaller width to make the trapezium rule estimate more accurate.
$\log_{10}\sqrt{x}=\dfrac12\log_{10}x$, so $\int_2^{14}\log_{10}\sqrt{x}\,dx\approx\dfrac12(10.10)=5.05$.
$\log_{10}(100x^3)=\log_{10}100+\log_{10}x^3=2+3\log_{10}x$.
$\int_2^{14}(2+3\log_{10}x)\,dx\approx2(12)+3(10.10)=54.30$.
$(a)\ 10.10.\quad (b)$ Use more strips. $\quad (c)(i)\ 5.05.\quad (c)(ii)\ 54.30$.
For $y=a^x+4$ with $a>1$, the curve is increasing.
At $x=0$, $y=a^0+4=5$, so the $y$-intercept is $(0,5)$.
As $x\to-\infty$, $a^x\to0$, so the horizontal asymptote is $y=4$.
The $x$ values increase by $0.3$, so $h=0.3$.
$\displaystyle \int_2^{3.5}(2^x-2x)dx\approx\dfrac{0.3}{2}[0+4.3137+2(0.3246+0.8629+1.6643+2.7896)]$.
This gives $2.339475$, so the estimate is $2.34$ to 2 decimal places.
$2^x+2x=(2^x-2x)+4x$, so $\displaystyle \int_2^{3.5}(2^x+2x)dx\approx2.34+\int_2^{3.5}4x\,dx$.
$\displaystyle \int_2^{3.5}4x\,dx=[2x^2]_2^{3.5}=16.5$, so the estimate is $2.34+16.5=18.84$.
$2^{x+1}-4x=2(2^x-2x)$, so the estimate is $2(2.34)=4.68$.
$(a)$ $y$-intercept $(0,5)$ and asymptote $y=4$. $\quad (b)$ $2.34$. $\quad (c)(i)$ $18.84$. $\quad (c)(ii)$ $4.68$.
For $y=a^{-x}+4$, set $x=0$: $y=a^0+4=5$, so the curve meets the $y$-axis at $(0,5)$.
Since $a>1$, $a^{-x}\to0$ as $x\to\infty$, so the horizontal asymptote is $y=4$.
The curve is decreasing, passes through $(0,5)$, and approaches the line $y=4$ from above.
The $x$ values increase by $2.5$, so $h=2.5$.
$\displaystyle \int_{-4}^{8.5}\left(3^{-x/2}+4\right)dx\approx \dfrac{2.5}{2}\left[13+4.009+2(6.280+4.577+4.146+4.037)\right]$.
This gives $68.86125$, so to two significant figures the estimate is $69$.
$\displaystyle \int_{-4}^{8.5}4\,dx=4(12.5)=50$.
Using the answer to part (b), $\displaystyle \int_{-4}^{8.5}3^{-x/2}dx\approx69-50=19$.
For $\displaystyle \int_{-8.5}^{4}(3^{x/2}+4)dx$, let $u=-x$. This changes it to the same integral as part (b).
Therefore the two integrals in part (c)(ii) are approximately equal, so the total is $69+69=138$.
$(a)$ $y$-intercept $(0,5)$ and asymptote $y=4$. $\quad (b)$ $69$. $\quad (c)(i)$ $19$. $\quad (c)(ii)$ $138$.
$y=\dfrac12x^2+1458x^{-3/2}-74$.
$\dfrac{dy}{dx}=x+1458\left(-\dfrac32\right)x^{-5/2}=x-2187x^{-5/2}$.
At a stationary point, $x-2187x^{-5/2}=0$.
$x=2187x^{-5/2}$, so $x^{7/2}=2187=3^7$.
Raising both sides to the power $\dfrac27$ gives $x=3^2=9$. Since $x>0$, this is the only stationary point.
At $x=9$, $y=\dfrac12(9)^2+\dfrac{1458}{\sqrt{9^3}}-74=\dfrac{81}{2}+54-74=\dfrac{41}{2}$.
The region lies from $x=4$ to $x=9$ between the curve and the horizontal line $y=\dfrac{41}{2}$.
$\displaystyle R=\int_4^9\left(\dfrac12x^2+1458x^{-3/2}-74-\dfrac{41}{2}\right)dx$.
$\displaystyle \int\left(\dfrac12x^2+1458x^{-3/2}-\dfrac{189}{2}\right)dx=\dfrac{x^3}{6}-2916x^{-1/2}-\dfrac{189}{2}x$.
$\displaystyle R=\left[\dfrac{x^3}{6}-2916x^{-1/2}-\dfrac{189}{2}x\right]_4^9=\dfrac{373}{3}$.
$(a)$ Proven. $\quad (b)$ $\dfrac{373}{3}$.
$y=\dfrac{9x-x^2}{2\sqrt{x}}=\dfrac12(9x^{1/2}-x^{3/2})$.
$\dfrac{dy}{dx}=\dfrac12\left(\dfrac92x^{-1/2}-\dfrac32x^{1/2}\right)=\dfrac{3(3-x)}{4\sqrt{x}}$.
For $x>0$, the denominator is not zero. So $\dfrac{dy}{dx}=0$ when $3-x=0$.
Therefore the $x$ coordinate of $P$ is $3$.
$y=0$ when $9x-x^2=0$, so $x=9$ for the positive intercept.
The region is under the curve from $x=1$ to $x=9$, so $R=\displaystyle \int_1^9 \dfrac{9x-x^2}{2\sqrt{x}}dx$.
$R=\displaystyle \int_1^9 \dfrac12(9x^{1/2}-x^{3/2})dx=\left[3x^{3/2}-\dfrac15x^{5/2}\right]_1^9$.
$R=\left(3(27)-\dfrac15(243)\right)-\left(3-\dfrac15\right)=\dfrac{148}{5}$.
$(a)$ $x=3$. $\quad (b)$ $\dfrac{148}{5}$.
$y=\dfrac{8\sqrt{x}-5}{2x^2}=4x^{-3/2}-\dfrac52x^{-2}$.
$\int\left(4x^{-3/2}-\dfrac52x^{-2}\right)dx=-8x^{-1/2}+\dfrac{5}{2x}$.
$\left[-8x^{-1/2}+\dfrac{5}{2x}\right]_2^4=\left(-4+\dfrac58\right)-\left(-4\sqrt2+\dfrac54\right)$.
This gives $4\sqrt2-\dfrac{37}{8}$.
$\int\left(\dfrac12x^2+k\right)dx=\dfrac{x^3}{6}+kx$.
$\left[\dfrac{x^3}{6}+kx\right]_{-3}^{6}=\left(36+6k\right)-\left(-\dfrac92-3k\right)=\dfrac{81}{2}+9k$.
$\dfrac{81}{2}+9k=55$, so $9k=\dfrac{29}{2}$ and $k=\dfrac{29}{18}$.
$(i)\ 4\sqrt2-\dfrac{37}{8}.\quad (ii)\ k=\dfrac{29}{18}$.
For $y=x^3-4x^{5/2}-kx^{1/2}+28x-44$, $\dfrac{dy}{dx}=3x^2-10x^{3/2}-\dfrac{k}{2}x^{-1/2}+28$.
$\dfrac{dy}{dx}=3x^2-10x^{3/2}-\dfrac{k}{2\sqrt{x}}+28$.
At $T$, $x=9$ and $\dfrac{dy}{dx}=0$.
$0=3(9)^2-10(9)^{3/2}-\dfrac{k}{2\sqrt9}+28=243-270-\dfrac{k}{6}+28$.
This simplifies to $0=1-\dfrac{k}{6}$, so $k=6$.
With $k=6$, the $y$ coordinate of $T$ is $9^3-4(9)^{5/2}-6\sqrt9+28(9)-44=-53$.
The line $NT$ is $y=-53$. The region is between the curve and this line from $x=0$ to $x=9$.
$R=\displaystyle\int_0^9\left(x^3-4x^{5/2}-6x^{1/2}+28x-44+53\right)dx$.
$R=\left[\dfrac{x^4}{4}-\dfrac87x^{7/2}-4x^{3/2}+14x^2+9x\right]_0^9$.
$R=\dfrac{6939}{28}=247.821\ldots$, so the area is $248$ to $3$ significant figures.
$(a)$ $3x^2-10x^{3/2}-\dfrac{k}{2\sqrt{x}}+28$. $\quad (b)$ Proven. $\quad (c)$ $248$.
$\dfrac{(x-k)^2}{\sqrt{x}}=\dfrac{x^2-2kx+k^2}{x^{1/2}}=x^{3/2}-2kx^{1/2}+k^2x^{-1/2}$.
$\displaystyle \int \left(x^{3/2}-2kx^{1/2}+k^2x^{-1/2}\right)dx=\dfrac25x^{5/2}-\dfrac{4k}{3}x^{3/2}+2k^2x^{1/2}$.
From $1$ to $16$, this gives $\left(\dfrac{2048}{5}-\dfrac{256k}{3}+8k^2\right)-\left(\dfrac25-\dfrac{4k}{3}+2k^2\right)$.
The result is $6k^2-84k+\dfrac{2046}{5}$, so $a=6$ and $b=-84$.
Since $A(1,9)$ lies on $C$, $9=\dfrac{(1-k)^2}{\sqrt1}=(1-k)^2$.
Thus $1-k=\pm3$, so $k=4$ or $k=-2$. Since $k$ is positive, $k=4$.
When $k=4$ and $x=16$, $q=\dfrac{(16-4)^2}{\sqrt{16}}=\dfrac{144}{4}=36$, so $B=(16,36)$.
The line through $A(1,9)$ and $B(16,36)$ has gradient $\dfrac{36-9}{16-1}=\dfrac95$, so $l$ has equation $y=\dfrac95x+\dfrac{36}{5}$.
$\displaystyle \int_1^{16}\left(\dfrac95x+\dfrac{36}{5}\right)dx=\left[\dfrac{9}{10}x^2+\dfrac{36}{5}x\right]_1^{16}=\dfrac{675}{2}$.
From part (a), with $k=4$, $\displaystyle \int_1^{16}\dfrac{(x-4)^2}{\sqrt{x}}dx=6(4)^2-84(4)+\dfrac{2046}{5}=\dfrac{846}{5}$.
The shaded area is $\dfrac{675}{2}-\dfrac{846}{5}=\dfrac{1683}{10}$.
$(a)$ $a=6,\ b=-84$. $\quad (b)$ $k=4$. $\quad (c)$ Area $=\dfrac{1683}{10}$.
At an intersection, $x-x^2=mx$, so $x(1-m-x)=0$.
The non-zero intersection has $x=1-m$, and then $y=m(1-m)$, so $P=(1-m,m(1-m))$.
Region $R_1$ is between the curve and the line from $x=0$ to $x=1-m$.
$A_{R_1}=\int_0^{1-m}\{(x-x^2)-mx\}\,dx=\int_0^{1-m}\{(1-m)x-x^2\}\,dx$.
$A_{R_1}=\left[\dfrac{1-m}{2}x^2-\dfrac13x^3\right]_0^{1-m}=\dfrac{(1-m)^3}{2}-\dfrac{(1-m)^3}{3}=\dfrac{(1-m)^3}{6}$.
The total area under $y=x-x^2$ from $x=0$ to $x=1$ is $\int_0^1(x-x^2)\,dx=\dfrac16$.
Since $R_1$ and $R_2$ are equal and together make this total area, each has area $\dfrac{1}{12}$.
$\dfrac{(1-m)^3}{6}=\dfrac1{12}$, so $(1-m)^3=\dfrac12$ and $m=1-\dfrac1{\sqrt[3]{2}}$.
$(a)\ P=(1-m,\ m(1-m)).\quad (b)$ Proven: $\dfrac{(1-m)^3}{6}$. $\quad (c)\ m=1-\dfrac1{\sqrt[3]{2}}$.
For $y=\dfrac43x^3-11x^2+kx$, $\dfrac{dy}{dx}=4x^2-22x+k$.
At the turning point $M$, $x=2$ and $\dfrac{dy}{dx}=0$.
$4(2)^2-22(2)+k=0$, so $16-44+k=0$ and $k=28$.
With $k=28$, $\dfrac{dy}{dx}=4x^2-22x+28=2(2x-7)(x-2)$.
The derivative is positive outside its roots, so $y$ is increasing for $x<2$ or $x>\dfrac72$.
At $x=2$, $y=\dfrac43(8)-11(4)+28(2)=\dfrac{68}{3}$. Hence line $l$ has equation $y=\dfrac{68}{3}$.
The region is between the line and the curve from $x=0$ to $x=2$, so $A=\int_0^2\left(\dfrac{68}{3}-\left(\dfrac43x^3-11x^2+28x\right)\right)dx$.
$A=\left[\dfrac{68}{3}x-\dfrac13x^4+\dfrac{11}{3}x^3-14x^2\right]_0^2$.
Substituting $x=2$ gives $\dfrac{136}{3}-\dfrac{16}{3}+\dfrac{88}{3}-56=\dfrac{40}{3}$.
$(a)$ Proven: $k=28$. $\quad (b)\ x<2\text{ or }x>\dfrac72.\quad (c)\ \dfrac{40}{3}$.
The $x$ values increase by $0.25$, so $h=0.25$.
The area under $C_1$ is approximately $\dfrac{0.25}{2}\{5.453+5+2(7.764+9.375+9.964+9.367+7.626)\}$.
This gives $12.330625$, so the estimate is $12.33$ to 2 decimal places.
$\int(x^{3/2}-3x+9)\,dx=\dfrac25x^{5/2}-\dfrac32x^2+9x+c$.
Between $x=2.5$ and $x=4$, the exact integral of $C_2$ is $\left[\dfrac25x^{5/2}-\dfrac32x^2+9x\right]_{2.5}^{4}=7.722\ldots$.
The shaded region is between $C_1$ and $C_2$, so its area is approximately $12.330625-7.722\ldots=4.608\ldots$.
To one decimal place, the area of region $R$ is $4.6$.
$(a)\ 12.33.\quad (b)\ \dfrac25x^{5/2}-\dfrac32x^2+9x+c.\quad (c)\ 4.6$.
The volume is $l^2h=250000$, so $l^2=\dfrac{250000}{h}$.
The open box has one square base and four side faces, so $S=l^2+4lh$.
Since $l=\sqrt{\dfrac{250000}{h}}=\dfrac{500}{\sqrt h}$, $4lh=4\cdot\dfrac{500}{\sqrt h}\cdot h=2000\sqrt h$.
Therefore $S=\dfrac{250000}{h}+2000\sqrt h$.
$S=250000h^{-1}+2000h^{1/2}$, so $\dfrac{dS}{dh}=-250000h^{-2}+1000h^{-1/2}$.
$-250000h^{-2}+1000h^{-1/2}=0$, so $1000h^{-1/2}=250000h^{-2}$.
Multiplying by $h^2$ gives $1000h^{3/2}=250000$, so $h^{3/2}=250$ and $h=250^{2/3}$.
Since $h=250^k$, we have $k=\dfrac23$.
$\dfrac{d^2S}{dh^2}=500000h^{-3}-500h^{-3/2}$.
At $h=250^{2/3}$, $h^{3/2}=250$, so the second derivative is $\dfrac{500000}{250^2}-\dfrac{500}{250}=8-2=6>0$.
The stationary point is therefore a minimum.
$(a)$ Proven: $S=\dfrac{250000}{h}+2000\sqrt h$. $\quad (b)\ h=250^{2/3}$, so $k=\dfrac23$. $\quad (c)$ Minimum.
From $(x-k)^2+(y-2k)^2=k+7$, the centre is $(k,2k)$ and the radius is $\sqrt{k+7}$.
Since $P(2,3)$ lies on the circle, $(2-k)^2+(3-2k)^2=k+7$.
Expanding gives $k^2-4k+4+4k^2-12k+9=k+7$, hence $5k^2-17k+6=0$.
$5k^2-17k+6=(5k-2)(k-3)$, so $k=\dfrac25$ or $k=3$.
For the tangent calculation, $k<2$, so the centre is $\left(\dfrac25,\dfrac45\right)$.
The radius from the centre to $P$ has gradient $\dfrac{3-4/5}{2-2/5}=\dfrac{11}{8}$, so the tangent gradient is $-\dfrac8{11}$.
The tangent is $y-3=-\dfrac8{11}(x-2)$. Setting $y=0$ gives $x=\dfrac{49}{8}$, so $T=\left(\dfrac{49}{8},0\right)$.
Triangle $OPT$ has base $\dfrac{49}{8}$ on the $x$-axis and height $3$, so its area is $\dfrac12\cdot\dfrac{49}{8}\cdot3=\dfrac{147}{16}$.
$(a)(i)\ (k,2k).\quad (a)(ii)\ \sqrt{k+7}.\quad (b)(ii)\ k=3\text{ or }\dfrac25.\quad (c)\ \dfrac{147}{16}$.
$y=\dfrac{9x^2(5-\sqrt{x})}{5}=9x^2-\dfrac95x^{5/2}$.
$\dfrac{dy}{dx}=18x-\dfrac92x^{3/2}=\dfrac92x(4-\sqrt{x})$.
For the turning point shown away from the origin, set $4-\sqrt{x}=0$. Hence $x=16$.
At $x=16$, $y=\dfrac{9(16)^2(5-4)}{5}=\dfrac{2304}{5}$.
Therefore $M\left(16,\dfrac{2304}{5}\right)$.
For the positive intercept, $\dfrac{9x^2(5-\sqrt{x})}{5}=0$ gives $5-\sqrt{x}=0$, so $x=25$.
The top boundary is $y=\dfrac{2304}{5}$ and the curve is below it from $x=16$ to $x=25$. So $R=\displaystyle\int_{16}^{25}\left(\dfrac{2304}{5}-9x^2+\dfrac95x^{5/2}\right)dx$.
$R=\left[\dfrac{2304}{5}x-3x^3+\dfrac{18}{35}x^{7/2}\right]_{16}^{25}$.
$R=\dfrac{9189}{7}=1312.714\ldots$, so the area is $1312.7$ to one decimal place.
$(a)$ $M\left(16,\dfrac{2304}{5}\right)$. $\quad (b)$ $x_P=25$. $\quad (c)$ $1312.7$.
For $y=\dfrac23x^2-9\sqrt{x}+13$, $\dfrac{dy}{dx}=\dfrac43x-\dfrac{9}{2\sqrt{x}}$.
$y$ is increasing when $\dfrac{dy}{dx}>0$.
For $x>0$, multiply by $6\sqrt{x}$: $8x\sqrt{x}-27>0$.
$x^{3/2}>\dfrac{27}{8}=\left(\dfrac32\right)^3$, so $\sqrt{x}>\dfrac32$ and $x>\dfrac94$.
At $P(9,40)$, $\dfrac{dy}{dx}=\dfrac43(9)-\dfrac{9}{2\cdot3}=12-\dfrac32=\dfrac{21}{2}$.
$y-40=\dfrac{21}{2}(x-9)$, so $y=\dfrac{21}{2}x-\dfrac{109}{2}$.
Set $y=0$: $\dfrac{21}{2}x-\dfrac{109}{2}=0$, so $x=\dfrac{109}{21}$.
From $x=0$ to $x=\dfrac{109}{21}$, the lower boundary is the $x$-axis. From $x=\dfrac{109}{21}$ to $x=9$, the lower boundary is the tangent.
$\displaystyle R=\int_0^{109/21}\left(\dfrac23x^2-9\sqrt{x}+13\right)dx+\int_{109/21}^{9}\left(\dfrac23x^2-9\sqrt{x}+13-\left(\dfrac{21}{2}x-\dfrac{109}{2}\right)\right)dx$.
$\displaystyle \int\left(\dfrac23x^2-9\sqrt{x}+13\right)dx=\dfrac{2}{9}x^3-6x^{3/2}+13x$, and $\displaystyle \int\left(\dfrac{21}{2}x-\dfrac{109}{2}\right)dx=\dfrac{21}{4}x^2-\dfrac{109}{2}x$.
Substituting the limits and simplifying gives $R=\dfrac{857}{21}$.
$(a)$ $y$ is increasing for $x>\dfrac94$. $\quad (b)$ Area $=\dfrac{857}{21}$.
For $y=x^3-5x^2+3x+14$, $\dfrac{dy}{dx}=3x^2-10x+3$.
$3x^2-10x+3=0$ factors as $(3x-1)(x-3)=0$, so $x=\dfrac13$ or $x=3$.
$\dfrac{d^2y}{dx^2}=6x-10$. At $x=3$, this is positive, so the minimum turning point has $x=3$.
Substitute $x=3$: $y=27-45+9+14=5$. Hence $T=(3,5)$.
At $x=2$ on $C_1$, $y=8-20+6+14=8$, so $A=(2,8)$.
The centre of $C_2$ is $T=(3,5)$ and it passes through $A=(2,8)$. Thus $r^2=(2-3)^2+(8-5)^2=10$.
Therefore $C_2$ has equation $(x-3)^2+(y-5)^2=10$.
The radius $TA$ has gradient $\dfrac{8-5}{2-3}=-3$, so the tangent at $A$ has gradient $\dfrac13$.
Using $A=(2,8)$, $y-8=\dfrac13(x-2)$, so $y=\dfrac13x+\dfrac{22}{3}$.
From $x=0$ to $x=2$, the cubic is above the line. The required area is $\displaystyle \int_0^2 \left[x^3-5x^2+3x+14-\left(\dfrac13x+\dfrac{22}{3}\right)\right]dx$.
The integrand is $x^3-5x^2+\dfrac83x+\dfrac{20}{3}$.
$\displaystyle \left[\dfrac{x^4}{4}-\dfrac{5x^3}{3}+\dfrac{4x^2}{3}+\dfrac{20x}{3}\right]_0^2=4-\dfrac{40}{3}+\dfrac{16}{3}+\dfrac{40}{3}=\dfrac{28}{3}$.
$(a)$ $T=(3,5)$. $\quad (b)$ $(x-3)^2+(y-5)^2=10$. $\quad (c)$ $y=\dfrac13x+\dfrac{22}{3}$. $\quad (d)$ $\dfrac{28}{3}$.
Elite Mathematics · Questions classified by topic, solutions checked by Dr Eslam
Elite Mathematics