Sec 4 A Math: Integration, practice questions & worked solutions

$$\begin{aligned} \int\left(x - \frac{5}{2x}\right)^2 dx &= \int\left(x^2 - 5 + \frac{25}{4}x^{-2}\right) dx\\ &= \frac{x^3}{3} - 5x - \frac{25}{4x} + c \end{aligned}$$

(a)

$$\begin{aligned} y &= e^x(1+\cos x)\\ \frac{dy}{dx} &= e^x(1+\cos x)+e^x(-\sin x)\\ \frac{dy}{dx} &= e^x(1+\cos x-\sin x) \end{aligned}$$

(b)

$$\begin{aligned} \int e^x(1+\cos x-\sin x)\,dx &= \int e^x+e^x(\cos x-\sin x)\,dx = e^x(1+\cos x)+c_1\\ \int e^x\,dx+\int e^x(\cos x-\sin x)\,dx &= e^x(1+\cos x)+c\\ e^x+\int e^x(\cos x-\sin x)\,dx &= e^x(1+\cos x)+c\\ \int e^x(\cos x-\sin x)\,dx &= e^x(1+\cos x)-e^x+c\\ \int e^x(\cos x-\sin x)\,dx &= e^x\cos x+c \end{aligned}$$

(a)

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left(2e^x + 1\right)^{-1} &= -\frac{2e^x}{\left(2e^x + 1\right)^2} \end{aligned}$$

(b)

$$\begin{aligned} \int -\frac{2e^x}{\left(2e^x + 1\right)^2}\, \mathrm{d}x &= \frac{1}{2e^x + 1}\ (+c)\\ \int \frac{e^x}{\left(2e^x + 1\right)^2}\, \mathrm{d}x &= \frac{-1}{4e^x + 2} + c \end{aligned}$$

$$\begin{aligned} f''(x) &= -2\cos x+12\sin 2x\\ f'(x) &= \int(-2\cos x+12\sin 2x)\,\mathrm{d}x\\ &= -2\sin x+\frac{12(-\cos 2x)}{2}+c\\ &= -2\sin x-6\cos 2x+c\\ f(x) &= \int(-2\sin x-6\cos 2x+c)\,\mathrm{d}x\\ &= -2(-\cos x)-\frac{6\sin 2x}{2}+cx+d\\ &= 2\cos x-3\sin 2x+cx+d \end{aligned}$$ At $(0,0)$, $0=2\cos 0-3\sin 2(0)+c(0)+d$: $$\begin{aligned} 2+d &= 0\\ d &= -2 \end{aligned}$$ At $\left(\dfrac{\pi}{2},-2\right)$, $-2=2\cos\dfrac{\pi}{2}-3\sin 2\!\left(\dfrac{\pi}{2}\right)+\dfrac{\pi}{2}c-2$: $$\begin{aligned} -3\sin\pi+\frac{\pi}{2}c &= 0\\ \frac{\pi}{2}c &= 0\\ c &= 0\\ f(x) &= 2\cos x-3\sin 2x-2\\ f(\pi) &= 2\cos\pi-3\sin 2\pi-2\\ &= 2(-1)-3(0)-2\\ &= -4 \end{aligned}$$

(a)

$$\begin{aligned} y &= \ln\left(\sin^2 x\right)\\ &= 2\ln(\sin x)\\ \frac{dy}{dx} &= 2\left(\frac{1}{\sin x}\right)(\cos x)\\ &= 2\cot x \end{aligned}$$

(b)

$$\begin{aligned} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\!\cot x - \cos^2 2x\, dx &= \frac{1}{2}\Bigl[2\ln\sin x\Bigr]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos 4x + 1}{2}\, dx\\ &= \frac{1}{2}\left[0 - 2\ln\frac{1}{\sqrt{2}}\right] - \frac{1}{2}\left[\frac{\sin 4x}{4} + x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\\ &= \frac{1}{2}\ln 2 - \frac{1}{2}\left[\frac{\pi}{2} - \frac{\pi}{4}\right]\\ &= \frac{1}{2}\ln 2 - \frac{\pi}{8}\\ &= -\frac{\pi}{8} + \frac{1}{2}\ln 2 \end{aligned}$$

(a)

When $y = \pi$, $\dfrac{5}{x} = \pi$,

Therefore, $x = \dfrac{5}{\pi}$

Point $P$ is $\left(\dfrac{5}{\pi},\, \pi\right)$.

When $x = \dfrac{3\pi}{2}$, $y = 5 \div \dfrac{3\pi}{2} = \dfrac{10}{3\pi}$

Point $Q$ is $\left(\dfrac{3\pi}{2},\, \dfrac{10}{3\pi}\right)$.

(b)

$$\begin{aligned} \text{Shaded area} &= \pi\left(\frac{5}{\pi}\right) + \int_{\frac{5}{\pi}}^{\frac{3\pi}{2}} \frac{5}{x}\, dx\\ &= 5 + 5\bigl[\ln x\bigr]_{\frac{5}{\pi}}^{\frac{3\pi}{2}}\\ &= 5 + 5\left[\ln\!\left(\frac{3\pi}{2}\right) - \ln\!\left(\frac{5}{\pi}\right)\right]\\ &= 10.4 \ \text{units}^2 \ (\text{3 s.f.}) \end{aligned}$$

(c)

$$\begin{aligned} \text{Area of big rectangle} &= \pi\left(\frac{3\pi}{2}\right)\\ &= \frac{3\pi^2}{2} \ \text{units}^2\\ \text{Area of small rectangle} &= \left(\frac{3\pi}{2}\right)\!\left(\frac{10}{3\pi}\right)\\ &= 5 \ \text{units}^2 \end{aligned}$$ Therefore, area of shaded region is between $5$ units$^2$ and $\dfrac{3\pi^2}{2}$ units$^2$.

(i)

$$\begin{aligned} \frac{11x + 12}{x^2(x + 4)} &= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 4}\\ 11x + 12 &= Ax(x + 4) + B(x + 4) + Cx^2 \end{aligned}$$ when $x = 0$, $$\begin{aligned} 12 &= 0 + 4B + 0\\ B &= 3 \end{aligned}$$ when $x = -4$, $$\begin{aligned} -32 &= 0 + 0 + 16C\\ C &= -2 \end{aligned}$$ when $x = 1$, $$\begin{aligned} 23 &= 5A + 15 - 2\\ A &= 2 \end{aligned}$$ $$\therefore \frac{11x + 12}{x^2(x + 4)} = \frac{2}{x} + \frac{3}{x^2} - \frac{2}{x + 4}$$

(ii)

$$\begin{aligned} \int \frac{11x + 12}{x^2(x + 4)}\, dx &= \int \frac{2}{x} + \frac{3}{x^2} - \frac{2}{x + 4}\, dx\\ &= 2\ln x - \frac{3}{x} - 2\ln(x + 4) + c \end{aligned}$$

(a)

$$\begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= -4\sin(2x+\pi)\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 2\cos(2x+\pi)+c_1\\ 3 &= 2\cos\!\left[2\!\left(\frac{\pi}{2}\right)+\pi\right]+c_1\\ 3 &= 2(1)+c_1\\ c_1 &= 1\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 2\cos(2x+\pi)+1\\ y &= \sin(2x+\pi)+x+c_2\\ \pi &= \sin\!\left[2\!\left(\frac{\pi}{2}\right)+\pi\right]+\frac{\pi}{2}+c_2\\ c_2 &= \frac{\pi}{2}\\ y &= \sin(2x+\pi)+x+\frac{\pi}{2} \end{aligned}$$

(b)

When $x=3$, $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2\cos[2(3)+\pi]+1=-0.920$.

$y$ is decreasing when $x=3$.

(a)

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= \int(3e^{2x}-e^{-x})\,\mathrm{d}x\\ &= \frac{3}{2}e^{2x}+e^{-x}+c \end{aligned}$$ Gradient $=4$ at $(0,1)$, $$\begin{aligned} 4 &= \frac{3}{2}+1+c\\ c &= 1.5\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3}{2}e^{2x}+e^{-x}+\frac{3}{2} \end{aligned}$$ since $\dfrac{3}{2}e^{2x}>0$, $e^{-x}>0$, $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3}{2}e^{2x}+e^{-x}+\frac{3}{2}>0,$$ the curve has no stationary points.

(b)

$$\begin{aligned} y &= \int\!\left(\frac{3}{2}e^{2x}+e^{-x}+\frac{3}{2}\right)\mathrm{d}x\\ &= \frac{3}{4}e^{2x}-e^{-x}+\frac{3}{2}x+d \end{aligned}$$ at $(0,1)$, $1=\dfrac{3}{4}-1+0+d$, $d=\dfrac{5}{4}$. $$y = \frac{3}{4}e^{2x}-e^{-x}+\frac{3}{2}x+\frac{5}{4}$$

(a)

$$\begin{aligned} \frac{1 - 3x - 3x^2}{x(x + 1)^2} &= \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}\\ 1 - 3x - 3x^2 &= A(x + 1)^2 + Bx(x + 1) + Cx \end{aligned}$$ $$\begin{aligned} \text{When } x &= -1, & C &= -1\\ \text{When } x &= 0, & A &= 1\\ \text{When } x &= 1, & B &= -4 \end{aligned}$$ $$\frac{1 - 3x - 3x^2}{x(x + 1)^2} = \frac{1}{x} - \frac{4}{x + 1} - \frac{1}{(x + 1)^2}$$

(b)

$$\begin{aligned} \int \frac{1 - 3x - 3x^2}{2x(x + 1)^2}\ dx &= \frac{1}{2}\int \frac{1 - 3x - 3x^2}{x(x + 1)^2}\ dx\\ &= \frac{1}{2}\int \frac{1}{x} - \frac{4}{x + 1} - \frac{1}{(x + 1)^2}\ dx\\ &= \frac{1}{2}\ln x - 2\ln(x + 1) + \frac{1}{2(x + 1)} + C \end{aligned}$$ Alternative forms: $\ln\sqrt{x} - \ln(x + 1)^2 + \dfrac{1}{2(x + 1)} + C$ $$\text{or } \ln\frac{\sqrt{x}}{(x + 1)^2} + \frac{1}{2(x + 1)} + C$$

(a)

$$\begin{aligned} x &= (y - 2)^2 \text{--- (1)}\\ y &= x - 10 \text{--- (2)} \end{aligned}$$ Sub (1) into (2): $$\begin{aligned} y &= (y - 2)^2 - 10\\ y &= (y^2 - 4y + 4) - 10\\ y^2 - 5y - 6 &= 0\\ (y + 1)(y - 6) &= 0\\ y &= -1 \text{ or } y = 6\\ x &= 9 \text{ or } x = 16 \end{aligned}$$ Coordinates of $M$ and $N$ are $(9, -1)$ and $(16, 6)$ respectively.

(b)

$$\begin{aligned} \text{area of region } A &= \frac{1}{2}(9 + 16)(6 + 1) - \int_{-1}^{6}(y - 2)^2\, dy\\ &= 87\frac{1}{2} - \left[\frac{(y - 2)^3}{3}\right]_{-1}^{6}\\ &= 87\frac{1}{2} - \left[\frac{(6 - 2)^3}{3} - \frac{(-1 - 2)^3}{3}\right]\\ &= 57\frac{1}{6}\text{ units}^2 \end{aligned}$$ $$\begin{aligned} \text{area of region } B &= \int_{-1}^{2}(y - 2)^2\, dy + \frac{1}{2}(10 - 1)(9)\\ &= \left[\frac{(y - 2)^3}{3}\right]_{-1}^{2} + \frac{81}{2}\\ &= \frac{(2 - 2)^3}{3} - \frac{(-1 - 2)^3}{3} + \frac{81}{2}\\ &= 49\frac{1}{2}\text{ units}^2 \end{aligned}$$

(a)

$$\begin{aligned} y &= -3(2x-4)^{-\frac{1}{3}}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= -3\!\left(-\frac{1}{3}\right)(2x-4)^{-\frac{4}{3}}(2)\\ &= \frac{2}{(2x-4)^{\frac{4}{3}}} \end{aligned}$$ When $x=-2$, $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2}{[2(-2)-4]^{\frac{4}{3}}}=\dfrac{1}{8}$.

Gradient of normal at $P=-8$. $$\begin{aligned} \text{Equation of normal at } P: y-1.5 &= -8(x+2)\\ y &= -8x-\frac{29}{2} \end{aligned}$$ At $Q$, $y=0$: $$0=-8x-\frac{29}{2} \;\Rightarrow\; x=-\frac{29}{16}$$ $$Q\!\left(-\frac{29}{16},\,0\right)$$

(b)

$$\begin{aligned} \text{Area of shaded region} &= \int_{-2}^{0} -3(2x-4)^{-\frac{1}{3}}\,\mathrm{d}x-\frac{1}{2}(2-1.8125)(1.5)\\ &= \left[\frac{-3(2x-4)^{\frac{2}{3}}}{\frac{2}{3}(2)}\right]_{-2}^{0}-\frac{9}{64}\\ &= \left[-\frac{9}{4}(2x-4)^{\frac{2}{3}}\right]_{-2}^{0}-\frac{9}{64}\\ &= -\frac{9}{4}\!\left[(-4)^{\frac{2}{3}}-(-8)^{\frac{2}{3}}\right]-\frac{9}{64}\\ &= 3.190\ \text{units}^2\ (\text{to 3 d.p.}) \end{aligned}$$