Use \(\tan^2(x/2)=\sec^2(x/2)-1\). Thus \[\int_0^{\pi/2}2\tan^2(x/2)\,dx=\left[4\tan(x/2)-2x\right]_0^{\pi/2}=4-\pi.\]

Use \(\cos^22x=(1+\cos4x)/2\) and \(\tan^2x=\sec^2x-1\): \[4\cos^22x+3\tan^2x=2\cos4x+3\sec^2x-1.\] Hence \[I=\left[\frac12\sin4x+3\tan x-x\right]_{\pi/6}^{\pi/3}.\]

At \(\pi/3\), the value is \(-\sqrt3/4+3\sqrt3-\pi/3\). At \(\pi/6\), it is \(\sqrt3/4+\sqrt3-\pi/6\). Therefore \[\boxed{I=\frac{3\sqrt3}{2}-\frac\pi6}.\]

(a) Three equal intervals give \(h=1\), and \[\begin{array}{c|rrrr}x&0&1&2&3\\\hline y&3&\sqrt{10}&\sqrt{17}&6\end{array}\]

The trapezium estimate is \[T=\frac12\left(3+2\sqrt{10}+2\sqrt{17}+6\right)=11.78538\ldots=\boxed{11.8}\quad\text{(3 s.f.)}.\]

(b) It is an overestimate: the curve is concave upwards, so the straight tops of the trapeziums lie above the curve.

Differentiate: \[\frac{dy}{dx}=4\sin2x\cos2x+2\sin2x=2\sin2x(2\cos2x+1).\] The two maxima satisfy \(\cos2x=-1/2\), giving \[\alpha=\frac\pi3,\qquad\beta=\frac{2\pi}3.\] The second derivative is \(-6\) at each of these points, confirming maxima.

Use \(\sin^22x=(1-\cos4x)/2\): \[y=\frac72-\frac12\cos4x-\cos2x.\] The curve is positive throughout the required region, so its area is \[A=\int_{\pi/3}^{2\pi/3}y\,dx=\left[\frac72x-\frac18\sin4x-\frac12\sin2x\right]_{\pi/3}^{2\pi/3}.\]

At the upper limit this is \(7\pi/3+3\sqrt3/16\); at the lower limit it is \(7\pi/6-3\sqrt3/16\). Hence \[\boxed{A=\frac{7\pi}{6}+\frac{3\sqrt3}{8}}.\]

First expand the integrand: \((1+e^{-x})^2=1+2e^{-x}+e^{-2x}\).

An antiderivative is \(F(x)=x-2e^{-x}-\frac12e^{-2x}\). Apply the limits \(a\) and \(2a\):

\[\begin{aligned}\int_a^{2a}(1+e^{-x})^2\,\mathrm{d}x&=\left[x-2e^{-x}-\frac12e^{-2x}\right]_a^{2a}\\&=2a-2e^{-2a}-\frac12e^{-4a}-\left(a-2e^{-a}-\frac12e^{-2a}\right)\\&=\boxed{a+2e^{-a}-\frac32e^{-2a}-\frac12e^{-4a}}.\end{aligned}\]

At the two intercepts bounding the shaded region, \[8\cos2x-\frac1{\cos^22x}=0\ \Longrightarrow\ 8\cos^32x=1\ \Longrightarrow\ \cos2x=\frac12.\]

These intercepts are \(x=-\pi/6\) and \(x=\pi/6\). The curve is above the axis between them. Since \(1/\cos^22x=\sec^22x\),

\[\begin{aligned}A&=\int_{-\pi/6}^{\pi/6}(8\cos2x-\sec^22x)\,\mathrm{d}x\\&=\left[4\sin2x-\frac12\tan2x\right]_{-\pi/6}^{\pi/6}\\&=2\left(4\frac{\sqrt3}2-\frac{\sqrt3}2\right)=\boxed{3\sqrt3}.\end{aligned}\]

Use \(\sin^2x=(1-\cos2x)/2\): \[\int6\sin^2x\,dx=\int(3-3\cos2x)\,dx=3x-\frac32\sin2x+C.\]

(a) At \(B\), \[8e^{-x/2}-1=0\Rightarrow e^{-x/2}=\frac18\Rightarrow\frac x2=\ln8=3\ln2.\] Thus \(x=6\ln2\).

(b) At \(x=0\), \(y=7\). The triangle under the chord has area \(\frac12(6\ln2)(7)=21\ln2\). The area under the curve is \[\int_0^{6\ln2}(8e^{-x/2}-1)\,dx=\left[-16e^{-x/2}-x\right]_0^{6\ln2}=14-6\ln2.\] Subtracting gives the shaded area \[21\ln2-(14-6\ln2)=27\ln2-14.\]

(a) \[\frac{dy}{dx}=-8\sin2x+8\cos x=8\cos x(1-2\sin x).\] Stationary points occur at \(x=\pi/6,\pi/2,5\pi/6\). Since \[\frac{d^2y}{dx^2}=-16\cos2x-8\sin x,\] the second derivative is \(-12\) at \(\pi/6\) and \(5\pi/6\), but \(8\) at \(\pi/2\). The two maxima have \(y=6\), so \[A=(\pi/6,6),\qquad B=(5\pi/6,6).\]

(b) The chord is \(y=6\), so \[\text{area}=\int_{\pi/6}^{5\pi/6}(6-4\cos2x-8\sin x)\,dx=\left[6x-2\sin2x+8\cos x\right]_{\pi/6}^{5\pi/6}=4\pi-6\sqrt3.\]

(a) \[\frac{dy}{dx}=12e^{2x}-3e^{3x}=3e^{2x}(4-e^x).\] This is zero at \(x=\ln4\); it changes from positive to negative there, so this is the maximum.

(b) The positive \(x\)-intercept satisfies \(e^{2x}(6-e^x)=0\), giving \(x=\ln6\). Hence \[\text{area}=\int_0^{\ln6}(6e^{2x}-e^{3x})\,dx=\left[3e^{2x}-\frac13e^{3x}\right]_0^{\ln6}=(108-72)-\left(3-\frac13\right)=\frac{100}{3}.\]

\[\int_2^{11}\frac8{4x+1}\,dx=\left[2\ln(4x+1)\right]_2^{11}=2\ln45-2\ln9=2\ln5=\ln25.\] Thus \(a=25\).

(a) Equate the curves: \[4e^{-2x}=1+0.5\sin3x\Rightarrow e^{-2x}=0.25+0.125\sin3x.\] Taking logarithms and dividing by \(-2\) gives the stated equation.

(b) Use \(x_{n+1}=-0.5\ln(0.25+0.125\sin3x_n)\), retaining full accuracy internally:

Hence \(x_P=0.4912\) to 4 significant figures.

(c) Between the \(y\)-axis and \(P\), the exponential is above the sine curve. Thus \[\text{area}=\int_0^{x_P}(4e^{-2x}-1-0.5\sin3x)\,dx=\left[-2e^{-2x}-x+\frac16\cos3x\right]_0^{x_P}\approx0.610\ldots.\] To 2 significant figures, the area is \(0.61\).

Integrate using the chain rule in reverse: \[y=8\tan(x/2)+C.\] Substituting \(x=\pi/2\), \(y=5\), gives \(5=8\tan(\pi/4)+C\), so \(C=-3\). The curve is \[y=8\tan(x/2)-3.\]

(a) At the intersection, \[e^{2x}=8e^{-x}\Rightarrow e^{3x}=8\Rightarrow3x=\ln8=3\ln2.\] Thus \(x=\ln2\).

(b) The upper curve between 0 and \(\ln2\) is \(8e^{-x}\). Hence \[\text{area}=\int_0^{\ln2}(8e^{-x}-e^{2x})\,dx=\left[-8e^{-x}-\frac12e^{2x}\right]_0^{\ln2}=(-4-2)-(-8-1/2)=\frac52.\]

(a) Division gives \[18x^3-6x^2-30x+4=(3x-1)(6x^2-10)-6.\] The quotient is \(6x^2-10\), and the remainder is \(-6\).

(b) The integrand is \(6x^2-10-6/(3x-1)\). Hence \[\begin{aligned}\int_1^5\left(6x^2-10-\frac6{3x-1}\right)dx&=\left[2x^3-10x-2\ln(3x-1)\right]_1^5\\&=208-2\ln7=208-\ln49.\end{aligned}\]

(a) By the chain rule, \(f^\prime(x)=\tan(x/2)\sec^2(x/2)\). Thus \(f^\prime(2\pi/3)=\sqrt3(4)=4\sqrt3\).

(b) Use \(\tan^2(x/2)=\sec^2(x/2)-1\): \[\int_0^{\pi/2}\left(\tan^2(x/2)+\sin x\right)\,dx=\left[2\tan(x/2)-x-\cos x\right]_0^{\pi/2}=3-\frac\pi2.\]

(a) \(f^\prime(x)=24\sin3x\cos3x=12\sin6x\). Thus \(f^\prime(\pi/4)=12\sin(3\pi/2)=-12\).

(b) \(4\sin^2(3x)=2-2\cos6x\), so \[\int f(x)\,dx=2x-\frac13\sin6x+C.\]

(a) \(h=1\), with ordinates \(\sqrt[3]7,\sqrt[3]{12},3\): \[A\approx\frac12\left(\sqrt[3]7+2\sqrt[3]{12}+3\right)=4.74589\ldots=4.75.\]

(b) \[A+B=\int_0^2\frac{27}{2x+5}\,dx=\left[\frac{27}{2}\ln(2x+5)\right]_0^2=\frac{27}{2}\ln\frac95.\]

(c) \(B\approx\frac{27}{2}\ln(9/5)-4.74589\ldots=3.19\).

(d) It is an underestimate: the tops of the trapezia lie above the cubic-root curve, so (a) overestimates A. Subtracting this from the exact total underestimates B.

(a) \(h=\pi/12\), with ordinates \(0,\sqrt{3/4},\sqrt{\sqrt3/2+3/4}\). \[A\approx\frac{\pi}{24}\left(2\sqrt{3/4}+\sqrt{\sqrt3/2+3/4}\right)=0.393129\ldots=0.39.\]

(b) \[V=\pi\int_0^{\pi/6}(\sin2x+\sin^22x)\,dx=\pi\int_0^{\pi/6}\left(\sin2x+\frac12-\frac12\cos4x\right)\,dx.\] Hence \[V=\pi\left[-\frac12\cos2x+\frac x2-\frac18\sin4x\right]_0^{\pi/6}=\frac\pi4+\frac{\pi^2}{12}-\frac{\pi\sqrt3}{16}.\]

(a) \[p(x)=(3x+2)(3x^2+4)+(k-8).\] Multiplying the quotient \(3x^2+4\) gives \(9x^3+6x^2+12x+8\), leaving remainder \(k-8\).

(b) \[\int_1^6\frac{p(x)}{3x+2}\,dx=\left[x^3+4x+\frac{k-8}{3}\ln(3x+2)\right]_1^6=235+\frac{k-8}{3}\ln4.\] Comparing with \(a+\ln64= a+3\ln4\), gives \(a=235\) and \((k-8)/3=3\), so \(k=17\).

(a) \(dy/dx=-8e^{-x}-2e^{2x}\). At A, \(x=0\), so the gradient is \(-10\).

(b) \(8e^{-x}=e^{2x}\Rightarrow e^{3x}=8\Rightarrow x=\ln2\). \[A=\int_0^{\ln2}(8e^{-x}-e^{2x})\,dx=\left[-8e^{-x}-\frac12e^{2x}\right]_0^{\ln2}=(-4-2)-(-8-\tfrac12)=\frac52.\]

(a) \[p(x)=(3x+2)(3x^2+4x-1)+6.\] The quotient is \(3x^2+4x-1\), remainder 6.

(b) \[\int_0^2\left(3x^2+4x-1+\frac6{3x+2}\right)\,dx=\left[x^3+2x^2-x+2\ln(3x+2)\right]_0^2=14+2\ln4=14+\ln16.\]

(a) \(h=2\), with ordinates \(\sqrt2,\sqrt{1+e},\sqrt{1+e^2},\sqrt{1+e^3}\). \[A\approx\frac22\left(\sqrt2+2\sqrt{1+e}+2\sqrt{1+e^2}+\sqrt{1+e^3}\right)=15.7.\]

(b) \[V=\pi\int_0^6y^2\,dx=\pi\int_0^6(1+e^{0.5x})\,dx=\pi[x+2e^{0.5x}]_0^6=4\pi+2\pi e^3.\]

(a) \[\int_4^{10}\frac4{2x-5}\,dx=[2\ln(2x-5)]_4^{10}=2\ln15-2\ln3=2\ln5=\ln25.\]

(b) \[\int_4^{10}e^{2x-5}\,dx=\left[\frac12e^{2x-5}\right]_4^{10}=\frac12(e^{15}-e^3).\]

(a) \(dy/dx=-3e^{-x/2}\), giving gradient \(-3/e\) at \(x=2\).

(b) \(A=(0,6)\), so the horizontal boundary is \(y=6\). The shaded area is \[\int_0^2(6-6e^{-x/2})\,dx=\left[6x+12e^{-x/2}\right]_0^2=\frac{12}{e}.\]

(a) Polynomial division gives \[6x^3-5x^2-24x-4=(2x+1)(3x^2-4x-10)+6.\] Therefore the quotient is \(3x^2-4x-10\) and the remainder is 6.

(b) The integrand is \(3x^2-4x-10+6/(2x+1)\). Hence \[\int_2^7\cdots\,dx=\left[x^3-2x^2-10x+3\ln(2x+1)\right]_2^7\] \[=195+3\ln(15/5)=195+\ln27.\]

Use \(4\cos^22x=2+2\cos4x\) and \(1/\cos^2x=\sec^2x\). Then \[\int_{\pi/4}^{\pi/3}(2+2\cos4x+\sec^2x)\,dx=\left[2x+\frac12\sin4x+\tan x\right]_{\pi/4}^{\pi/3}\] \[=\left(\frac{2\pi}{3}-\frac{\sqrt3}{4}+\sqrt3\right)-\left(\frac{\pi}{2}+1\right)=\frac{3\sqrt3}{4}+\frac{\pi}{6}-1.\]

The curve meets \(y=2\) at \(x=0\). Therefore \[\text{area}=\int_0^6\left(2-\frac6{2x+3}\right)\,dx=\left[2x-3\ln(2x+3)\right]_0^6\] \[=12-3\ln(15/3)=12-\ln125.\]

(a) The first quotient term is \(12x^3/(4x)=3x^2\). Subtracting \(3x^2(4x-3)\) leaves \(8x-4\). The next quotient term is 2; subtracting \(2(4x-3)=8x-6\) leaves 2. Thus \[p(x)=(4x-3)(3x^2+2)+2,\] with quotient \(3x^2+2\) and remainder 2.

(b) The integrand simplifies to \(2+2/(4x-3)\), and \(4x-3>0\) throughout the interval. Hence \[\int_2^{12}\left(2+\frac2{4x-3}\right)\,dx=\left[2x+\frac12\ln(4x-3)\right]_2^{12}=20+\frac12\ln\frac{45}{5}=20+\ln3.\]

For \(0\le x\le4\), \(6/(3x+2)>0\), while \(3e^{-x}-3\le0\). Thus \[A=\int_0^4\left(\frac6{3x+2}-3e^{-x}+3\right)\,dx.\] Integrating, \[A=\left[2\ln(3x+2)+3e^{-x}+3x\right]_0^4.\] Therefore \[A=2\ln14+3e^{-4}+12-2\ln2-3=2\ln7+9+3e^{-4}=\boxed{\ln49+9+3e^{-4}}.\]

(a) At an x-axis crossing, \[3\sin x-3\sin2x=3\sin x(1-2\cos x)=0.\] The interior crossing has \(\cos x=1/2\), giving \[a=\pi/3.\]

(b) The shaded region is the positive part from \(\pi/3\) to \(\pi\), so \[A=\int_{\pi/3}^{\pi}(3\sin x-3\sin2x)\,dx=\left[-3\cos x+\frac32\cos2x\right]_{\pi/3}^{\pi}.\] This is \[\left(3+\frac32\right)-\left(-\frac32-\frac34\right)=\frac{27}4.\]

(a) The first quotient term is \(2x^2\); subtracting \(2x^2(x+2)\) leaves \(x^2+ax+2a\). The next term is \(x\); subtracting \(x(x+2)\) leaves \((a-2)x+2a\). The next term is \(a-2\), leaving remainder 4. Thus \[p(x)=(x+2)(2x^2+x+a-2)+4.\] The quotient is \(2x^2+x+a-2\).

(b) Since \(x+2>0\) over the interval, \[\int_{-1}^1\frac{p(x)}{x+2}\,dx=\left[\frac23x^3+\frac12x^2+(a-2)x+4\ln(x+2)\right]_{-1}^1=2a-\frac83+4\ln3.\] Comparing with the given form gives \[2a-\frac83=\frac{22}3\quad\Longrightarrow\quad a=5,\qquad \ln b=4\ln3=\ln81\quad\Longrightarrow\quad b=81.\]

(a) With \(h=(4-1)/3=1\), the ordinates are \(\ln1,\ln2,\ln3,\ln4\). The trapezium approximation is \[T=\frac12(\ln1+2\ln2+2\ln3+\ln4)=\ln2+\ln3+\frac12\ln4=\ln2+\ln3+\ln2=\ln12.\]

(b) The logarithm graph is concave down for \(x>0\), since \(d^2y/dx^2=-1/x^2<0\). Each straight chord forming the top of a trapezium lies below the curve, so the three trapezia give an under-estimate.

The shaded area is \[A=\int_0^\pi\left(\sqrt{2\pi-2x}-\sin^2x\right)\,dx.\] For the first term, \[\int\sqrt{2\pi-2x}\,dx=-\frac13(2\pi-2x)^{3/2}.\] For the second, use \(\sin^2x=(1-\cos2x)/2\): \[\int\sin^2x\,dx=\frac x2-\frac14\sin2x.\] Thus \[A=\left[-\frac13(2\pi-2x)^{3/2}-\frac x2+\frac14\sin2x\right]_0^\pi=\boxed{\frac13(2\pi)^{3/2}-\frac\pi2}.\]

(a) By the chain rule, \[\frac{dy}{dx}=2\tan x\sec^2x=2\tan x(1+\tan^2x)=2\tan x+2\tan^3x.\]

(b) Part (a) gives \(\int(\tan x+\tan^3x)\,dx=\tfrac12\tan^2x\). Also \(\tan^2x=\sec^2x-1\), so \(\int\tan^2x\,dx=\tan x-x\). Therefore \[I=\left[\frac12\tan^2x+\tan x-x\right]_{\pi/4}^{\pi/3}=\left(\frac32+\sqrt3-\frac\pi3\right)-\left(\frac12+1-\frac\pi4\right)=\boxed{\sqrt3-\frac\pi{12}}.\]

Integrate each exponential using the chain rule in reverse: \[\int(4e^{2x}-2e^{-x})\,dx=2e^{2x}+2e^{-x}+C.\] Applying the limits gives \[\left[2e^{2x}+2e^{-x}\right]_{-1}^{2}=(2e^4+2e^{-2})-(2e^{-2}+2e)=\boxed{2e^4-2e}.\]

Because \(a>0\), all logarithms use positive arguments. Integrate and apply the limits: \[\frac13[\ln x]_a^{a+14}=\frac13\ln\frac{a+14}{a}=\ln2.\] Multiply by 3 and exponentiate: \[\ln\frac{a+14}{a}=3\ln2=\ln8,\qquad\frac{a+14}{a}=8.\] Hence \(a+14=8a\), so \[\boxed{a=2}.\]

(a) Integrate the exponential and use the limits: \[\left[3e^{2x+1}\right]_0^2=\boxed{3e^5-3e}.\]

(b) Use \(\tan^2x=\sec^2x-1\) and \(4\sin^2 2x=2-2\cos4x\). Thus the integrand becomes \(\sec^2x+1-2\cos4x\), so \[\boxed{\tan x+x-\frac12\sin4x+C}.\]

(a) Expand and use \(\sec x\cos x=1\): \[(\sec x+\cos x)^2=\sec^2x+2+\cos^2x\] \[=\sec^2x+2+\frac{1+\cos2x}{2}=\sec^2x+\frac52+\frac12\cos2x.\] Therefore \(\boxed{a=5/2,\quad b=1/2}\).

(b) Integrate the expression from (a): \[\left[\tan x+\frac52x+\frac14\sin2x\right]_0^{\pi/4}=1+\frac{5\pi}8+\frac14=\boxed{\frac54+\frac{5\pi}8}.\]

(a) The interval width is \(h=(4-1)/3=1\). Evaluate the four ordinates:

Using unrounded values, \[T=\frac12\left(3+2\frac6{1+\sqrt2}+2\frac6{1+\sqrt3}+2\right)=7.181433\ldots.\] Hence the approximation is \(\boxed{7.1814}\).

(b) Integrate exactly: \[\left[4e^{x/2-2}\right]_1^4=\boxed{4-4e^{-3/2}}.\]

(c) Subtract the integral of the lower curve from the approximation for the upper curve: \[A\approx7.181433\ldots-(4-4e^{-3/2})=4.073954\ldots,\qquad\boxed{A\approx4.07}.\]

(d) This is an over-estimate. The curve \(6/(1+\sqrt x)\) is concave upwards on the interval, so each trapezium lies above it. The lower curve was integrated exactly, so the area inherits that over-estimate.

(a) By the quotient rule, \[\boxed{\frac{dy}{dx}=\frac{5(4x^3+1)-60x^3}{(4x^3+1)^2}=\frac{5(1-8x^3)}{(4x^3+1)^2}}.\] Set the numerator equal to zero: \(8x^3=1\), so \(\boxed{x=1/2}\). The numerator changes from positive to negative at this value, confirming a maximum.

(b) With \(h=(3-1)/2=1\), the ordinates are \(y(1)=1\), \(y(2)=10/33\) and \(y(3)=15/109\). Therefore \[A\approx\frac12\left(1+2\frac{10}{33}+\frac{15}{109}\right)=0.871837\ldots,\qquad\boxed{A\approx0.87}.\]

(c) This is an over-estimate: on the interval from 1 to 3 the curve is concave upwards, so the top of each trapezium lies above the curve.