The graph of y = x² - 6x + 10 cuts the y-axis at A. The graphs of y = x² - 6x + 10 and y = x + 10 cut one another at A and B. The line BC is perpendicular to the x-axis. Calculate the area of the shaded region enclosed by the curve and the line AB, showing all your working. [Figure 8.1]

Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

1. Method 1 (Separate areas subtracted):
2. [B1] [XB = XC =] 7 soi
3. [M2] Integral of (x² - 6x + 10) dx = x³/3 - 3x² + 10x or M1 for at least one term correct
4. [DM1] Correct or correct ft substitution of limits 0 and their 7 into their integral which must be at least two terms; condone omission of lower limit;
5. [B2] Area of trapezium = 1/2 (10+17)×7 oe or M1 for 1/2 (their 10 + their 17) × their 7 oe or B1 for Integral of (x+10) dx = x²/2 + 10x
6. [M1] dep on a genuine attempt to integrate the equation of the curve; must be their area trapezium/under the line - their attempt at area under curve
7. [A1] 343/6 or 57 1/6 or 57.2 to 3 sf or 57.16(6...) rot to 4 figs isw from full and correct working with no omitted steps
8. Method 2 (Subtracting and using integration once):
9. [B1] [XB = XC =] 7 soi
10. [B1] Integral of (-x² + 7x)dx condone omission of dx
11. [M3] Integral of (-x² + 7x)dx = -x³/3 + 7x²/2 oe or M2 for Integral of (px² + qx)dx = px³/3 + qx²/2 oe either with p = -1 or q = 7 or M1 for Integral of (px² + qx)dx = px³/3 + qx²/2 with non-zero constants p and q, with p ≠ -1 and q ≠ -7
12. [M2] Correct or correct ft substitution of limits 0 and their 7 dep on a valid integration attempt; evidence of substitution must be seen; condone omission of lower limit;
13. [A1] 343/6 or 57 1/6 or 57.2 to 3 sf or 57.16(6...) rot to 4 figs isw from full and correct working with no omitted steps