Further calculus – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: an improper integral has an infinite limit or an integrand that is unbounded at an end point. Replace the problem limit by a letter (\(R\) or \(a\)), integrate normally, then let the letter tend to its limit. If the limit is finite, the integral converges. Mark scheme as shown in brackets.

   1. Replace \(\infty\) by \(R\) and integrate: \(\int6e^{-2x}dx=-3e^{-2x}\). \(\int_0^R6e^{-2x}dx=3\left(1-e^{-2R}\right)\) [1] \(\to\) 3 [1] So the integral converges to 3.
   2. Replace \(\infty\) by \(R\) and integrate \(x^{-1/2}\). \(\int_1^R x^{-1/2}dx=2\sqrt R-2\) [1], which tends to infinity as \(R\to\infty\), so the integral diverges [1]

2. Answer to Question 2 (5 marks)

   Method: use \(\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+c\) and \(\displaystyle\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+c\). If the quadratic is not in this form, complete the square first and then substitute \(u=x+k\). Mark scheme as shown in brackets.

   1. Complete the square, then use the standard result with \(a=1\) and \(u=x+1\). \(x^2+2x+2=(x+1)^2+1\) [1]. \(\left[\arctan(x+1)\right]_0^1\) [1] \(=\arctan2-\arctan1=\arctan2-\dfrac\pi4\) [1] Numerically this is about 0.322.
   2. Here \(a=2\). \(\left[\arcsin\dfrac x2\right]_0^1\) [1] \(=\arcsin\dfrac12=\) \(\dfrac\pi6\) [1]

3. Answer to Question 3 (5 marks)

   1. \(V=\pi\displaystyle\int_0^1\dfrac1{1+x^2}dx\) [1] \(=\pi\left[\arctan x\right]_0^1\) [1] \(=\pi\cdot\dfrac\pi4=\dfrac{\pi^2}4\) [1]
   2. At \(x=1\), \(y^2=\tfrac12\), so the area is \(\dfrac\pi2\) [1]. A cylinder has volume \(\dfrac\pi2\approx1.57\), which is less than \(V\approx2.47\), since the radius is larger for \(x\lt 1\) [1]

   Method: for rotation about the \(x\)-axis, \(V=\pi\int y^2\,dx\), and here \(y^2=\dfrac1{1+x^2}\), a standard arctan integral. Common errors: forgetting the factor \(\pi\), or integrating \(y\) instead of \(y^2\). In (b), the comparison works because \(y\) decreases from 1 at \(x=0\) to \(\tfrac1{\sqrt2}\) at \(x=1\), so the cylinder fits inside the solid.

4. Answer to Question 4 (3 marks)

   Method: the mean value of \(f\) on \([a,b]\) is \(\dfrac1{b-a}\displaystyle\int_a^bf(x)\,dx\): the height of the rectangle on \([a,b]\) with the same area as the region under the curve. Mark scheme as shown in brackets.

   1. \(\int_0^4(4x-x^2)dx=\left[2x^2-\dfrac{x^3}3\right]_0^4\) [1] \(=32-\dfrac{64}3=\dfrac{32}3\) [1]. Mean \(=\dfrac14\cdot\dfrac{32}3=\) \(\dfrac83\) [1]

   Answer: the mean value is \(\tfrac83\approx2.67\), which is \(\tfrac23\) of the maximum value 4 of the parabola.

   By symmetry about \(x=2\), the same mean value is obtained on \([0,2]\).

5. Answer to Question 5 (3 marks)

   Method: rotating about the \(x\)-axis gives \(V=\pi\displaystyle\int_a^by^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^dx^2\,dy\), with the limits in \(y\). Square the function first, then integrate. Mark scheme as shown in brackets.

   1. The top of the planter is where \(y=9\). \(9=\dfrac{x^2}4\), \(x=6\): 6 cm [1]
   2. Rotating about the \(y\)-axis, write \(x^2=4y\) and integrate with respect to \(y\) from 0 to 9. \(x^2=4y\), \(V=\pi\displaystyle\int_0^94y\,dy\) [1] \(=\pi\cdot2\cdot81=162\pi\approx\) 509 cm\(^3\) [1] So the planter holds about 509 cm³ of compost.

Total : 20 marks