Integration – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Write each term as a power of \(x\) and integrate term by term; in (b) expand the bracket first.

   (a) \(2x^4+3x^{-1}+\tfrac23x^{3/2}+c\) (2). (b) Expand \(x^2-6x+9\): \(\dfrac{x^3}3-3x^2+9x+c\) (2).

   Check (a) by differentiating: \(8x^3-3x^{-2}+x^{1/2}\) ✓. Do not forget the constant \(c\) in each answer.

   Answer (b) can be checked by differentiating: \(x^2-6x+9=(x-3)^2\) ✓.

   In (a), \(-3x^{-2}\) integrates to \(-3\times\dfrac{x^{-1}}{-1}=3x^{-1}\), so the sign changes.

2. Answer to Question 2 (4 marks)

   Integrate the gradient function term by term, then substitute the point \((2,5)\) to find the constant.

   \(y=x^3-4x^2+x+c\) (2). \(8-16+2+c=5\), so \(c=11\) (1). \(y=x^3-4x^2+x+11\) (1).

   Check: at \(x=2\), \(8-16+2+11=5\) ✓. So the curve is \(y=x^3-4x^2+x+11\).

   Check by differentiating: \(\dfrac{dy}{dx}=3x^2-8x+1\) ✓.

   Mark scheme: 2 for the integral (1 if one term is wrong), 1 for using \((2,5)\), 1 for the final equation. Without the constant the curve would pass through \((2,-6)\), not \((2,5)\).

3. Answer to Question 3 (4 marks)

   Write \(\dfrac6{\sqrt x}=6x^{-1/2}\), which integrates to \(12x^{1/2}\); then substitute the limits 4 and 1.

   \(\left[12x^{1/2}+\dfrac{x^2}2\right]_1^4\) (2) \(=(24+8)-(12+0.5)\) (1) \(=19.5\) (1).

   So the value of the integral is 19.5.

   Note that \(\sqrt4=2\), so \(12x^{1/2}\) equals 24 at \(x=4\) and 12 at \(x=1\).

   A frequent error is integrating \(6x^{-1/2}\) to \(3x^{1/2}\) by multiplying by the power instead of dividing by it. The numerical integration function of a calculator also gives 19.5.

4. Answer to Question 4 (4 marks)

   Factorise to find the roots, which are the limits; the curve is below the axis between them.

   Roots \(x=2,4\) (1). \(\int_2^4(x^2-6x+8)dx=\left[\tfrac{x^3}3-3x^2+8x\right]_2^4\) (1) \(=\left(\tfrac{64}3-48+32\right)-\left(\tfrac83-12+16\right)=\tfrac{16}3-\tfrac{20}3=-\tfrac43\) (1). Area \(=\tfrac43\) (1).

   The integral is negative because the region is below the \(x\)-axis, so the area is its modulus, \(\tfrac43\) square units.

   By symmetry, the region lies between \(x=2\) and \(x=4\), with the lowest point of the curve at \((3,-1)\).

5. Answer to Question 5 (4 marks)

   1. Compare the heights of the line and the curve at a point between the intersections. At \(x=2\): line 2, curve 1 (1) Since \(2\gt 1\), the line is above the curve on the whole interval.
   2. Integrate (line \(-\) curve) from 0 to 4. \(\int_0^4\left(x-\tfrac14x^2\right)dx=\left[\tfrac{x^2}2-\tfrac{x^3}{12}\right]_0^4\) (1) \(=8-\tfrac{16}3=\tfrac83\) (2 for evaluation) The area between the ramp and the line is \(\tfrac83\approx2.67\) m\(^2\).

   Integrating curve minus line would give \(-\tfrac83\); an area is always positive.

Total : 20 marks