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2020 Paper 2 Q8
D: 1500.0 B: 1500.0
integration quartic polynomial curve sketching roots of polynomials area between curves symmetry inflection points

In this question, \(\mathrm{f}(x)\) is a quartic polynomial where the coefficient of \(x^4\) is equal to \(1\), and which has four real roots, \(0\), \(a\), \(b\) and \(c\), where \(0 < a < b < c\). \(\mathrm{F}(x)\) is defined by \(\mathrm{F}(x) = \displaystyle\int_0^x \mathrm{f}(t)\,\mathrm{d}t\). The area enclosed by the curve \(y = \mathrm{f}(x)\) and the \(x\)-axis between \(0\) and \(a\) is equal to that between \(b\) and \(c\), and half that between \(a\) and \(b\).

  1. Sketch the curve \(y = \mathrm{F}(x)\), showing the \(x\) co-ordinates of its turning points. Explain why \(\mathrm{F}(x)\) must have the form \(\mathrm{F}(x) = \frac{1}{5}x^2(x-c)^2(x-h)\), where \(0 < h < c\). Find, in factorised form, an expression for \(\mathrm{F}(x) + \mathrm{F}(c-x)\) in terms of \(c\), \(h\) and \(x\).
  2. If \(0 \leqslant x \leqslant c\), explain why \(\mathrm{F}(b) + \mathrm{F}(x) \geqslant 0\) and why \(\mathrm{F}(b) + \mathrm{F}(x) > 0\) if \(x \neq a\). Hence show that \(c - b = a\) or \(c > 2h\). By considering also \(\mathrm{F}(a) + \mathrm{F}(x)\), show that \(c = a + b\) and that \(c = 2h\).
  3. Find an expression for \(\mathrm{f}(x)\) in terms of \(c\) and \(x\) only. Show that the points of inflection on \(y = \mathrm{f}(x)\) lie on the \(x\)-axis.

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2019 Paper 2 Q2
D: 1500.0 B: 1500.0
integration inverse function geometric proof curve sketching implicit differentiation definite integral area between curves

The function f satisfies \(f(0) = 0\) and \(f'(t) > 0\) for \(t > 0\). Show by means of a sketch that, for \(x > 0\), $$\int_0^x f(t) \, dt + \int_0^{f(x)} f^{-1}(y) \, dy = xf(x).$$

  1. The (real) function g is defined, for all \(t\), by $$(g(t))^3 + g(t) = t.$$ Prove that \(g(0) = 0\), and that \(g'(t) > 0\) for all \(t\). Evaluate \(\int_0^2 g(t) \, dt\).
  2. The (real) function h is defined, for all \(t\), by $$(h(t))^3 + h(t) = t + 2.$$ Evaluate \(\int_0^8 h(t) \, dt\).

Solution:

TikZ diagram
Notice the total area is \(xf(x)\) and it is made up of the sum of the two integrals.
  1. Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\). \begin{align*} && t &= (g(t))^3 + g(t) \\ \Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\ \Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0 \end{align*}
    TikZ diagram
    From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
  2. \(\,\)
    TikZ diagram
    From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}

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2018 Paper 1 Q1
D: 1516.0 B: 1516.0
polynomials curve sketching tangent-normal integration area between curves inequality quadratic/cubic intersections triangle area

The line \(y=a^2 x\) and the curve \(y=x(b-x)^2\), where \(0 < a < b\,\), intersect at the origin \(O\) and at points \(P\) and \(Q \). The \(x\)-coordinate of \(P\) is less than the \(x\)-coordinate of \(Q\). Find the coordinates of \(P\) and \(Q\), and sketch the line and the curve on the same axes. Show that the equation of the tangent to the curve at \(P\) is \[ y = a(3a-2b)x + 2a(b-a)^2 . \] This tangent meets the \(y\)-axis at \(R\). The area of the region between the curve and the line segment \(OP\) is denoted by \(S\). Show that \[ S= \frac1{12}(b-a)^3(3a+b)\,. \] The area of triangle \(OPR\) is denoted by \(T\). Show that \(S>\frac{1}{3}T\,\).

Solution:

TikZ diagram
\begin{align*} && a^2x &= x(b-x)^2 \\ \Rightarrow && 0 &= x((b-x)^2-a^2) \\ &&&= x(b-a-x)(b+a-x)\\ && y &= x(b-x)^2 \\ \Rightarrow && y' &= (b-x)^2-2x(b-x) \\ P(b-a,a^2(b-a)): &&y' &= (b-(b-a))^2-2(b-a)(b-(b-a)) \\ &&&= a^2-2a(b-a) = a(3a-2b) \\ \Rightarrow && y &= a(3a-2b)(x-(b-a)) + a^2(b-a) \\ &&&= a(3a-2b)x + (b-a)(a^2-3a^2+2ba) \\ &&&= a(3a-2b)x + (b-a)2a(b-a) \\ &&&= a(3a-2b)x + 2a(b-a)^2 \\ \end{align*} Therefore the tangent at \(P\) is \(a(3a-2b)x + 2a(b-a)^2\) The area between the curve and \(OP\) is \begin{align*} &&S &= \int_0^{b-a} \left (x(b-x)^2-a^2x \right) \d x\\ &&&= \left [\frac{x^2}{2}b^2 - \frac{2x^3}{3}b +\frac{x^4}{4} - \frac{a^2x^2}{2}\right]_0^{b-a} \\ &&&= (b-a)^2 \tfrac12 (b^2-a^2) - \tfrac23(b-a)^3b + \tfrac14(b-a)^4 \\ &&&= \tfrac1{12}(b-a)^3(6(b+a)-8b+3(b-a)) \\ &&&= \tfrac1{12}(b-a)^3(b+3a) \end{align*} The area \([OPR] = T= \tfrac12 \cdot (b-a) \cdot 2a(b-a)^2 = a(b-a)^3\) Clearly \(S > \frac4{12}(b-a)^3a = \frac13T\)

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2014 Paper 1 Q2
D: 1500.0 B: 1484.0
integration integration by parts curve sketching logarithm absolute value area between curves asymptotes definite integral

  1. Show that \(\int \ln (2-x) \d x = -(2-x)\ln (2-x) + (2-x) + c \,,\ \) where \(x<2\).
  2. Sketch the curve \(A\) given by \(y= \ln \vert x^2-4\vert\).
  3. Show that the area of the finite region enclosed by the positive \(x\)-axis, the \(y\)-axis and the curve \(A\) is \(4\ln(2+\sqrt3)-2\sqrt3\,\).
  4. The curve \(B\) is given by \(y= \vert \ln \vert x^2-4\vert \vert\,\). Find the area between the curve \(B\) and the \(x\)-axis with \(| x| <2\). Note: you may assume that \(t \ln t \to 0\) as \(t\to 0\).

Solution:

  1. \(\,\) \begin{align*} && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\ && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\ &&&= -(2-x) \ln (2-x) +(2-x) + C \end{align*}
  2. \(\,\)
    TikZ diagram
  3. \begin{align*} && \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\ &&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\ &&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\ &&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\ &&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\ &&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\ &&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\ &&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3} \end{align*}
  4. TikZ diagram
    \begin{align*} && \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\ &&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3}) \end{align*}

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2010 Paper 2 Q8
D: 1600.0 B: 1482.0
integration exponential function trigonometric curve sketching geometric series area between curves summation integration by parts

The curves \(C_1\) and \(C_2\) are defined by \[ y= \e^{-x} \quad (x>0) \quad \text{ and } \quad y= \e^{-x}\sin x \quad (x>0), \] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram. Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0 < x_1 < x_2 < \cdots\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[ A_n = \tfrac12(\e^{2\pi}-1) \e^{-(4n+1)\pi/2} \] and hence find \(\displaystyle \sum_{n=1}^\infty A_n\).

Solution:

TikZ diagram
The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

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2008 Paper 1 Q6
D: 1516.0 B: 1500.0
integration inverse function exponential function curve sketching geometric proof area between curves definite integral

The function \(\f\) is defined by \[ \f(x) = \frac{\e^x-1}{\e-1}, \ \ \ \ \ x\ge0, \] and the function \(\g\) is the inverse function to \(\f\), so that \(\g(\f(x))=x\). Sketch \(\f(x)\) and \(\g(x)\) on the same axes. Verify, by evaluating each integral, that \[ \int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e +1)}\,, \] where \(\displaystyle k= \frac 1{\sqrt\e+1}\), and explain this result by means of a diagram.

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2007 Paper 2 Q2
D: 1600.0 B: 1500.0
curve sketching cubic stationary points point of inflection symmetry integration area between curves polynomial optimisation

A curve has equation \(y=2x^3-bx^2+cx\). It has a maximum point at \((p,m)\) and a minimum point at \((q,n)\) where \(p>0\) and \(n>0\). Let \(R\) be the region enclosed by the curve, the line \(x=p\) and the line \(y=n\).

  1. Express \(b\) and \(c\) in terms of \(p\) and \(q\).
  2. Sketch the curve. Mark on your sketch the point of inflection and shade the region \(R\). Describe the symmetry of the curve.
  3. Show that \(m-n=(q-p)^3\).
  4. Show that the area of \(R\) is \(\frac12 (q-p)^4\).

Solution:

  1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
  2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
    TikZ diagram
  3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
  4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

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2006 Paper 1 Q7
D: 1516.0 B: 1503.2
integration curve sketching absolute value trigonometric exponential area between curves cosec roots of equations modulus function

  1. Sketch on the same axes the functions \({\rm cosec}\, x\) and \(2x/ \pi\), for \(0 < x < \pi\,\). Deduce that the equation \(x\sin x = \pi/2 \) has exactly two roots in the interval \(0 < x < \pi\,\). Show that \[ \displaystyle \int_{\pi/2}^{\pi} \left \vert x\sin x - \frac{\pi} { 2} \right \vert \; \mathrm{d}x = 2\sin\alpha +\frac{3\pi^2} 4 - \alpha \pi -\pi -2\alpha \cos\alpha -1 \] where \(\alpha\) is the larger of the roots referred to above.
  2. Show that the region bounded by the positive \(x\)-axis, the \(y\)-axis and the curve \[y = \Bigl| \vert \e^x - 1 \vert - 1 \Bigr|\] has area \(\ln 4-1\).

Solution:

  1. \(\,\)
    TikZ diagram
    Notice that they are equal at \(1\) when \(x = \pi/2\), but this is a local minimum for \(\csc x\) whereas \(2x/\pi\) is increasing so there is a second intersection. Notice that \(\csc x = \frac{2x}{\pi} \Leftrightarrow x \sin x = \frac{\pi}{2}\) therefore our intersections are also the roots of \(x \sin x = \frac{\pi}{2}\) and the larger one is greater than \(\pi/2\) \begin{align*} && I &= \int_{\pi/2}^{\pi} \Bigl| x \sin x - \frac{\pi}{2} \Bigr| \d x \\ &&&= \int_{\pi/2}^{\alpha} \left ( x \sin x - \frac{\pi}{2} \right )\d x +\int_{\alpha}^{\pi} \left ( \frac{\pi}{2} -x \sin x \right) \d x \\ &&&= \left ( \pi - 2\alpha + \frac{\pi}{2}\right) \frac{\pi}{2} + \int_{\pi/2}^{\alpha} x \sin x\d x -\int_{\alpha}^{\pi} x \sin x \d x \\ &&&= \frac{3\pi^2}{4} - \alpha \pi + \left [-x \cos x \right]_{\pi/2}^{\alpha}+\left[x \cos x \right]_{\alpha}^{\pi} + \int_{\pi/2}^{\alpha} \cos x \d x - \int_{\alpha}^{\pi} \cos x \d x \\ &&&= \frac{3\pi^2}{4} - \alpha \pi -\alpha \cos \alpha -\pi -\alpha \cos \alpha+ \sin \alpha - 1+\sin \alpha \\ &&&= 2\sin \alpha + \frac{3\pi^2}{4} - \alpha \pi - 2\alpha \cos \alpha - 1 \end{align*}
  2. \(\,\)
    TikZ diagram
    \begin{align*} && A &= \int_0^{\ln 2} ||e^x-1|-1| \d x \\ &&&= \int_0^{\ln 2} |e^x-2| \d x \\ &&&=\int_0^{\ln 2} (2-e^x) \d x \\ &&&= 2 \ln 2 - \left [e^x \right]_0^{\ln 2} \\ &&&= \ln 4 - (2-1) = \ln 4 - 1 \end{align*}

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2006 Paper 2 Q5
D: 1600.0 B: 1469.6
curve sketching floor function integration area between curves piecewise functions summation quadratic

The notation \({\lfloor } x \rfloor\) denotes the greatest integer less than or equal to the real number \(x\). Thus, for example, \(\lfloor \pi\rfloor =3\,\), \(\lfloor 18\rfloor =18\,\) and \(\lfloor-4.2\rfloor = -5\,\).

  1. Two curves are given by \(y= x^2+3x-1\) and \(y=x^2 +3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Find the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer.
  2. Two curves are given by \(y= x^2+3x-1\) and \(y=\lfloor x\rfloor ^2+3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Show that the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer, is \[ \tfrac 16 (n-1)(3n+11)\,. \]

Solution:

  1. \(\,\)
    TikZ diagram
    The difference between the curves is \(3x - 3\lfloor x \rfloor\), which has area \(\frac32\) for each step. Therefore the area between the curves from \(1 \leq x \leq n\) is \(\frac32 (n-1)\)
  2. \(\,\)
    TikZ diagram
    The area between the curves is \(x^2 - \lfloor x \rfloor ^2 + 3(x - \lfloor x \rfloor)\). Looking at \begin{align*} && A &= \int_1^n \left ( x^2 - \lfloor x \rfloor ^2 \right )\d x \\ &&&= \frac{n^3-1^3}{3} - \sum_{k=1}^{n-1} k^2 \\ &&&= \frac{(n-1)(n^2+n+1)}{3} - \frac{(n-1)n(2n-1)}{6} \\ &&&= \frac{(n-1) \left (2n^2+2n+2-2n^2+n \right)}{6} \\ &&&= \frac{(n-1)(3n+2)}{6} \end{align*} Therefore the total area is \(\frac{(n-1)(3n+2)}{6}+\frac32(n-1) = \frac{(n-1)}{6}\left ( 3n+2+9\right) =\frac{(n-1)(3n+11)}{6}\)

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2001 Paper 3 Q2
D: 1700.0 B: 1484.0
integration inverse hyperbolic functions cosh inverse logarithmic form area between curves hyperbola ellipse definite integral

Show that \( \cosh^{-1} x = \ln ( x + \sqrt{x^2-1})\). Show that the area of the region defined by the inequalities \(\displaystyle y^2 \ge x^2-8\) and \(\displaystyle x^2\ge 25y^2 -16 \) is \((72/5) \ln 2\).

Solution: \begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}

TikZ diagram
\begin{align*} && A &= 4 \left ( \int_0^3 \frac15\sqrt{16+x^2} \d x - \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x \right) \\ \\ x = 4\sinh u: && \int_0^3 \sqrt{4^2+x^2} \d x &= \int_{u=0}^{u=\sinh^{-1}(3/4)} \sqrt{4^2 (1+\sinh^2 u)} 4 \cosh u \d u \\ &&&= \int_0^{\sinh^{-1}(3/4)} 16 \cosh^{2}u \d u \\ &&&= 8\int_0^{\sinh^{-1}(3/4)} (1+\cosh 2u) \d u \\ &&&= 8 \left[u + \frac12 \sinh 2u\right]_0^{\sinh^{-1}(3/4)} \\ &&&= 8 \left (\sinh^{-1}(3/4) + \frac12 \sinh \left ( 2 \sinh^{-1}(3/4) \right) \right) \\ \\ && \sinh^{-1}(3/4) &= \ln\left ( \frac34 + \sqrt{\left ( \frac{3}{4} \right)^2 + 1} \right) \\ &&&= \ln \left ( \frac34 +\frac{5}{4} \right) \\ &&&= \ln 2 \\ \\ \Rightarrow && \int_0^3 \sqrt{4^2+x^2} \d x &= 8 \ln 2 + 4 \left ( \frac{e^{2 \ln 2} - e^{-2\ln2}}{2} \right) \\ &&&= 8 \ln 2 + 2 \cdot 4 - 2\cdot \frac{1}{4} \\ &&&= 8 \ln 2 + \frac{15}{2} \end{align*} \begin{align*} x = 2\sqrt{2} \cosh u: && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= \int_{u=0}^{u = \cosh^{-1} \frac{3}{2\sqrt{2}}} \sqrt{8(\cosh^2 u - 1)} 2 \sqrt{2} \sinh u \d u \\ &&&= \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 8\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 2\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \cosh 2 u -1 \d u \\ &&&= 4 \left [\frac12 \sinh 2u - u \right]_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \\ \\ && \cosh^{-1} \frac{3}{2\sqrt{2}} &= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\left ( \frac{3}{2\sqrt{2}} \right)^2-1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{9}{8} - 1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{1}{8} } \right) \\ &&&= \ln \frac{4}{2\sqrt{2}} \\ &&&= \frac12 \ln 2 \\ \\ && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= 4 \left ( \frac12 \frac{e^{\ln2} - e^{-\ln2}}{2} - \frac12 \ln 2\right) \\ &&&= 2 - \frac12 -2 \ln 2 \\ &&&= \frac32 - 2 \ln 2 \end{align*} \begin{align*} A &= 4 \left (\frac15\left(8\ln 2 + \frac{15}2 \right)- \left ( \frac32 - 2 \ln 2\right)\right) \\ &=4\cdot \left( \frac{8}{5} + 2 \right) \ln 2 \\ &= \frac{72}{5} \ln 2 \end{align*}

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