Mathematics II

Take the integral over $[0,1]$: $\displaystyle\int_0^1\frac{dx}{1+x^2}$, with three points $x_0=0,\ x_1=0.5,\ x_2=1$, so $n=2$ and $h=0.5$.

Trapezoidal approximation

$T=\frac h2\left[f_0+2f_1+f_2\right]=\frac{0.5}{2}\left[1+1.6+0.5\right]=0.775$

Actual value: $\left[\tan^{-1}x\right]_0^1=\frac\pi4=0.7854$.

The actual error is $0.7854-0.775\approx0.0104$.

Error formula. For the composite trapezoidal rule,

$E=-\frac{(b-a)}{12}h^2f''(\xi),\qquad \xi\in(a,b)$

Here $f''(x)=\dfrac{6x^2-2}{(1+x^2)^3}$. Its derivative $f'''(x)=\dfrac{24x(1-x^2)}{(1+x^2)^4}\ge0$ on $[0,1]$, so $f''$ increases from $f''(0)=-2$ to $f''(1)=0.5$. Therefore $\max|f''|=2$.

$|E|\le\frac{1\cdot(0.5)^2}{12}\cdot2=\frac{0.5}{12}\approx0.0417$

The actual error $0.0104$ is below the bound $0.0417$, which verifies the bound numerically.

A function is continuous at $x=1$ if $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$.

Left-hand limit ($x<1$): $\lim_{x\to1^-}(3x^2+2)=3+2=5$

Right-hand limit ($x>1$): $\lim_{x\to1^+}(2x+3)=2+3=5$

So $\lim_{x\to1}f(x)=5$ exists. But $f(1)=4\neq 5$.

Hence $f(x)$ is not continuous at $x=1$. The discontinuity is removable, since redefining $f(1)=5$ would make the function continuous.

Let $f(x)=\sin 3x$. By definition, $f'(x)=\lim_{h\to0}\frac{\sin 3(x+h)-\sin 3x}{h}$

Using $\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$:

$$f'(x)=\lim_{h\to0}\frac{2\cos\left(3x+\frac{3h}{2}\right)\sin\frac{3h}{2}}{h} =\lim_{h\to0}3\cos\left(3x+\frac{3h}{2}\right)\cdot\frac{\sin\frac{3h}{2}}{\frac{3h}{2}}$$

As $h\to0$, $\frac{\sin(3h/2)}{3h/2}\to1$ and $\cos\left(3x+\frac{3h}{2}\right)\to\cos3x$. Therefore $\frac{d}{dx}(\sin 3x)=3\cos 3x$

The limit has the form $\frac{\infty}{\infty}$, so L'Hospital's rule applies.

$\lim_{x\to\infty}\frac{2x^2+3x}{1+5x^2}=\lim_{x\to\infty}\frac{4x+3}{10x}$

This is again $\frac{\infty}{\infty}$, so apply the rule once more: $=\lim_{x\to\infty}\frac{4}{10}=\frac{2}{5}$

Here $P=1-3Q$ is the price, so

Revenue: $R=PQ=Q-3Q^2$

Cost: $C=Q^2-2Q$

Profit: $\pi=R-C=Q-3Q^2-Q^2+2Q=3Q-4Q^2$

For maximum profit, $\frac{d\pi}{dQ}=3-8Q=0$, giving $Q=\frac{3}{8}$.

Second derivative: $\frac{d^2\pi}{dQ^2}=-8<0$, so profit is a maximum.

Output for maximum profit is $Q=\dfrac{3}{8}$ units. The maximum profit is $\pi=\frac{9}{8}-\frac{9}{16}=\frac{9}{16}$.

(a) Multiply numerator and denominator by $1+\sin x$: $\int\frac{dx}{1-\sin x}=\int\frac{1+\sin x}{1-\sin^2x}\,dx=\int\frac{1+\sin x}{\cos^2x}\,dx=\int(\sec^2x+\sec x\tan x)\,dx$ $=\tan x+\sec x+C$

(b) $\int_0^1(x^2+5)\,dx=\left[\frac{x^3}{3}+5x\right]_0^1=\frac13+5=\frac{16}{3}$

Factorise: $\dfrac{dy}{dx}=\dfrac{y(x+1)}{x(y+1)}$.

Separate the variables: $\frac{y+1}{y}\,dy=\frac{x+1}{x}\,dx\;\Rightarrow\;\left(1+\frac1y\right)dy=\left(1+\frac1x\right)dx$

Integrate: $y+\ln y=x+\ln x+C$

or equivalently $y-x+\ln\dfrac{y}{x}=C$, where $C$ is an arbitrary constant.

Augmented matrix: $[A|B]=\begin{bmatrix}1&1&-1&1\\2&3&3&3\\1&-3&3&2\end{bmatrix}$

Apply $R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$: $\begin{bmatrix}1&1&-1&1\\0&1&5&1\\0&-4&4&1\end{bmatrix}$

Apply $R_3\to R_3+4R_2$: $\begin{bmatrix}1&1&-1&1\\0&1&5&1\\0&0&24&5\end{bmatrix}$

Here $\operatorname{rank}(A)=\operatorname{rank}([A|B])=3=$ number of unknowns, so the system is consistent with a unique solution.

Back-substitution: $24x_3=5\Rightarrow x_3=\frac{5}{24}$ $x_2=1-5x_3=1-\frac{25}{24}=-\frac{1}{24}$ $x_1=1-x_2+x_3=1+\frac1{24}+\frac5{24}=\frac54$

Solution: $x_1=\dfrac54,\;x_2=-\dfrac1{24},\;x_3=\dfrac5{24}$.

Homogeneous equations. A system $AX=O$ in which every constant term on the right-hand side is zero is called a homogeneous system. It always has the trivial solution $X=O$, and has non-trivial solutions only if $|A|=0$.

Solution by inverse matrix. The system is $AX=B$ with $A=\begin{bmatrix}-2&2&1\\-8&7&-4\\9&-8&5\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}-4\\-47\\55\end{bmatrix}$

Determinant: $|A|=-2(35-32)-2(-40+36)+1(64-63)=-6+8+1=3\neq0$

so $A^{-1}$ exists.

Cofactors: $C_{11}=3,\;C_{12}=4,\;C_{13}=1$ $C_{21}=-18,\;C_{22}=-19,\;C_{23}=2$ $C_{31}=-15,\;C_{32}=-16,\;C_{33}=2$

Adjoint (transpose of the cofactor matrix): $\operatorname{adj}A=\begin{bmatrix}3&-18&-15\\4&-19&-16\\1&2&2\end{bmatrix}$

$A^{-1}=\frac13\begin{bmatrix}3&-18&-15\\4&-19&-16\\1&2&2\end{bmatrix}$

Then $X=A^{-1}B$:

$$X=\frac13\begin{bmatrix}3(-4)-18(-47)-15(55)\\4(-4)-19(-47)-16(55)\\-4+2(-47)+2(55)\end{bmatrix} =\frac13\begin{bmatrix}-12+846-825\\-16+893-880\\-4-94+110\end{bmatrix} =\frac13\begin{bmatrix}9\\-3\\12\end{bmatrix}$$

$x=3,\quad y=-1,\quad z=4$

Check in the first equation: $-6-2+4=-4$. In the third: $27+8+20=55$.

Rolle's theorem. If $f$ is (i) continuous on $[a,b]$, (ii) differentiable on $(a,b)$, and (iii) $f(a)=f(b)$, then there is at least one point $c\in(a,b)$ with $f'(c)=0$.

Geometric meaning. If the graph of a continuous, smooth curve has the same height at both end points $A(a,f(a))$ and $B(b,f(b))$, then somewhere between them the curve has a horizontal tangent (parallel to the $x$-axis, and to the chord $AB$).

Verification for $f(x)=x^2-4$ on $[-3,3]$.

1. $f$ is a polynomial, so it is continuous on $[-3,3]$.
2. $f'(x)=2x$ exists everywhere, so $f$ is differentiable on $(-3,3)$.
3. $f(-3)=9-4=5$ and $f(3)=9-4=5$, so $f(-3)=f(3)$.

All conditions hold. Now $f'(c)=2c=0\Rightarrow c=0$, and $0\in(-3,3)$.

Hence Rolle's theorem is verified, with $c=0$.

Here $a=0,\;b=2,\;n=4$, so $h=\dfrac{2-0}{4}=0.5$, and $f(x)=2x^2-1$.

Composite trapezoidal rule: $T=\frac h2\left[f_0+f_4+2(f_1+f_2+f_3)\right]=\frac{0.5}{2}\left[-1+7+2(-0.5+1+3.5)\right]$ $=0.25\,[6+8]=3.5$

Exact value: $\int_0^2(2x^2-1)\,dx=\left[\frac{2x^3}{3}-x\right]_0^2=\frac{16}{3}-2=\frac{10}{3}\approx3.3333$

Absolute error: $\left|3.5-\frac{10}{3}\right|=\frac16\approx0.1667$

Taking the definition as $f(x)=2x+3$ for $-\frac32\le x<\frac32$ and $f(x)=2x-3$ for $x\ge\frac32$.

Continuity at $x=0$. The point $0$ lies in the interval where $f(x)=2x+3$, a polynomial.

$f(0)=3,\qquad \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=2(0)+3=3$

Since $\lim_{x\to0}f(x)=f(0)=3$, $f$ is continuous at $x=0$.

Discontinuity at $x=\frac32$.

$\lim_{x\to\frac32^-}f(x)=2\left(\tfrac32\right)+3=6,\qquad \lim_{x\to\frac32^+}f(x)=2\left(\tfrac32\right)-3=0$

The left-hand and right-hand limits are not equal ($6\ne0$), so $\lim_{x\to 3/2}f(x)$ does not exist. Hence $f(x)$ is discontinuous at $x=\frac32$ (a jump discontinuity).

Differentiate both with respect to $t$:

$\frac{dx}{dt}=a(1+\cos t),\qquad \frac{dy}{dt}=a\sin t$

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\sin t}{a(1+\cos t)}=\frac{2\sin\frac t2\cos\frac t2}{2\cos^2\frac t2}=\tan\frac t2$

So $\dfrac{dy}{dx}=\tan\dfrac t2$.

L'Hospital's Rule. If $f(x)$ and $g(x)$ are differentiable near $a$ and $\lim_{x\to a}\frac{f(x)}{g(x)}$ takes the form $\frac00$ or $\frac\infty\infty$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ provided the right-hand limit exists.

Evaluation. At $x=0$ the numerator is $0\cdot1-\ln1=0$ and the denominator is $0$, so the form is $\frac00$.

$\lim_{x\to0}\frac{xe^x-\ln(1+x)}{x^2}=\lim_{x\to0}\frac{e^x+xe^x-\frac1{1+x}}{2x}$

This is again $\frac00$. Apply the rule once more:

$=\lim_{x\to0}\frac{2e^x+xe^x+\frac1{(1+x)^2}}{2}=\frac{2+0+1}{2}=\frac32$

Rolle's Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then there is at least one $c\in(a,b)$ with $f'(c)=0$.

Verification for $f(x)=x^2-9$ on $[-3,3]$.

1. $f$ is a polynomial, so it is continuous on $[-3,3]$ and differentiable on $(-3,3)$.
2. $f(-3)=9-9=0$ and $f(3)=9-9=0$, so $f(-3)=f(3)$.

All conditions hold. Now $f'(x)=2x=0$ gives $c=0$, and $0\in(-3,3)$.

Hence Rolle's Theorem is verified, with $c=0$.

(a) Using $\displaystyle\int\frac{dx}{x^2-a^2}=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|$ with $a=5$ (the integrand is continuous on $[0,2]$ since $x^2-25\neq0$ there):

$\int_0^2\frac{dx}{x^2-25}=\frac1{10}\left[\ln\left|\frac{x-5}{x+5}\right|\right]_0^2=\frac1{10}\left(\ln\frac37-\ln1\right)=\frac1{10}\ln\frac37\approx-0.0847$

(b) Reading the integrand as $\dfrac{2x+5}{\sqrt{x^2+5x}}$, put $u=x^2+5x$, so $du=(2x+5)\,dx$:

$\int\frac{2x+5}{\sqrt{x^2+5x}}\,dx=\int u^{-1/2}\,du=2\sqrt u+C=2\sqrt{x^2+5x}+C$

The left side is an exact derivative:

$\sin x\frac{dy}{dx}+y\cos x=\frac{d}{dx}(y\sin x)=-\sin x\cos x$

Integrate both sides:

$y\sin x=-\int\sin x\cos x\,dx=-\frac{\sin^2x}{2}+C$

$\boxed{y=-\frac{\sin x}{2}+\frac{C}{\sin x}}$

The system is rearranged so that the diagonal terms dominate:

$x=\frac{8+y-z}{4},\qquad y=\frac{1-3x-2z}{6},\qquad z=\frac{2-x+y}{3}$

Start with $x=y=z=0$ and always use the latest values.

The values have settled to four decimals, so

$x\approx1.872,\qquad y\approx-0.705,\qquad z\approx-0.192$

(The exact solution is $x=\frac{73}{39},\ y=-\frac{55}{78},\ z=-\frac5{26}$.)

Maximize $Z=7x_1+5x_2$ subject to $x_1+2x_2\le6$, $4x_1+3x_2\le6$, $x_1,x_2\ge0$.

Add slack variables $s_1,s_2$:

$x_1+2x_2+s_1=6,\qquad 4x_1+3x_2+s_2=6,\qquad Z-7x_1-5x_2=0$

Initial tableau

The most negative entry is $-7$, so $x_1$ enters; the minimum ratio is in row $s_2$, so $s_2$ leaves (pivot 4).

Second tableau ($R_2\to R_2/4$, $R_1\to R_1-R_2$, $Z\to Z+7R_2$)

All entries in the $Z$ row are non-negative, so the solution is optimal.

$x_1=\frac32,\quad x_2=0,\quad Z_{\max}=\frac{21}{2}=10.5$

Stationary points. A point $x=c$ in the domain of $f$ where $f'(c)=0$ is a stationary point (the tangent is horizontal). It may be a local maximum, a local minimum, or neither.

Inflection point. A point where the curve changes its concavity, i.e. $f''$ changes sign. At such a point $f''(c)=0$ (when $f''$ exists), and the graph crosses its tangent.

Two numbers with sum 20 and minimum sum of squares. Let the numbers be $x$ and $20-x$, and

$S(x)=x^2+(20-x)^2$

$S'(x)=2x-2(20-x)=4x-40=0\ \Rightarrow\ x=10$

$S''(x)=4>0$

so $S$ is a minimum at $x=10$. The numbers are $10$ and $10$, and the minimum sum of squares is $10^2+10^2=200$.

Take 3 points $x_0=0,\ x_1=0.5,\ x_2=1$, so $n=2$ and $h=0.5$.

$\int_0^1\frac{dx}{1+x}\approx\frac h3\left[f_0+4f_1+f_2\right]=\frac{0.5}{3}\left[1+2.66667+0.5\right]=0.69444$

Error. The exact value is $\ln2=0.69315$, so the actual error is $0.69444-0.69315\approx0.0013$.

The error bound is $|E|\le\frac{(b-a)h^4}{180}\max|f^{(4)}(x)|,\qquad f^{(4)}(x)=\frac{24}{(1+x)^5},\quad \max_{[0,1]}|f^{(4)}|=24$

$|E|\le\frac{24h^4}{180}=\frac{2h^4}{15}=\frac{2(0.0625)}{15}\approx0.0083$

The actual error ($0.0013$) is within this bound.

Points needed for accuracy $10^{-5}$.

$\frac{2h^4}{15}\le10^{-5}\ \Rightarrow\ h^4\le7.5\times10^{-5}\ \Rightarrow\ h\le0.0931$

$n=\frac1h\ge10.74$

Simpson's rule needs an even $n$, so $n=12$ (check: $h=\frac1{12}$ gives $E\le6.4\times10^{-6}$). That is $12$ subintervals, i.e. $13$ points.

At $x=\theta$ the numerator is $\theta\cos\theta-\theta\cos\theta=0$ and the denominator is $0$, so the form is $\frac00$. Differentiate numerator and denominator with respect to $x$ ($\theta$ is a constant):

$\lim_{x\to\theta}\frac{x\cos\theta-\theta\cos x}{x-\theta}=\lim_{x\to\theta}\frac{\cos\theta+\theta\sin x}{1}=\cos\theta+\theta\sin\theta$

By definition, $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ with $f(x)=e^{3x-1}$:

$f'(x)=\lim_{h\to0}\frac{e^{3x+3h-1}-e^{3x-1}}{h}=e^{3x-1}\lim_{h\to0}\frac{e^{3h}-1}{h}$

Using $\lim_{t\to0}\frac{e^{t}-1}{t}=1$ with $t=3h$:

$\lim_{h\to0}\frac{e^{3h}-1}{h}=3\lim_{h\to0}\frac{e^{3h}-1}{3h}=3$

$\frac{dy}{dx}=3e^{3x-1}$

Verification. $f(x)=\sin x$ is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$, and $f(0)=\sin0=0=\sin\pi=f(\pi)$. All the conditions of Rolle's theorem hold.

$f'(x)=\cos x=0\ \Rightarrow\ x=\frac\pi2\in(0,\pi)$

so $c=\frac\pi2$ exists and Rolle's theorem is verified.

Point of the curve. The tangent is parallel to the $x$-axis where $f'(c)=0$, i.e. at $c=\frac\pi2$, where $y=\sin\frac\pi2=1$. The point is $\left(\dfrac\pi2,\,1\right)$.

$f(x)=x^3+6x^2+9x-2$

$f'(x)=3x^2+12x+9=3(x+1)(x+3)=0\ \Rightarrow\ x=-3,\ -1$

$f''(x)=6x+12$

• At $x=-3$: $f''(-3)=-6<0$, so there is a maximum: $f(-3)=-27+54-27-2=-2$.
• At $x=-1$: $f''(-1)=6>0$, so there is a minimum: $f(-1)=-1+6-9-2=-6$.

Point of inflection. $f''(x)=0$ gives $x=-2$, and $f''$ changes sign from negative to positive there. Also $f(-2)=-8+24-18-2=-4$.

Hence the maximum value is $-2$ (at $x=-3$), the minimum value is $-6$ (at $x=-1$), and the inflection point is $(-2,-4)$.

(a) Let $u=x^2+5x$, so $du=(2x+5)\,dx$:

$\int\frac{2x+5}{\sqrt{x^2+5x}}\,dx=\int u^{-1/2}du=2\sqrt{x^2+5x}+C$

(b) Integrate by parts with $u=\log x$, $dv=e^{2x}dx$ (so $v=\frac{e^{2x}}2$):

$\int e^{2x}\log x\,dx=\frac{e^{2x}\log x}{2}-\frac12\int\frac{e^{2x}}{x}\,dx$

The remaining integral is not an elementary function; it is the exponential integral, $\int\frac{e^{2x}}{x}dx=\mathrm{Ei}(2x)$. Hence

$\int e^{2x}\log x\,dx=\frac{e^{2x}\log x}{2}-\frac12\,\mathrm{Ei}(2x)+C$