Date | May 2019 | Marks available | 3 | Reference code | 19M.1.SL.TZ2.S_10 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 2 |
Command term | Find | Question number | S_10 | Adapted from | N/A |
Question
Let y=(x3+x)32.
Consider the functions f(x)=√x3+x and g(x)=6−3x2√x3+x, for x ≥ 0.
The graphs of f and g are shown in the following diagram.
The shaded region R is enclosed by the graphs of f, g, the y-axis and x=1.
Find dydx.
Hence find ∫(3x2+1)√x3+xdx.
Write down an expression for the area of R.
Hence find the exact area of R.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of choosing chain rule (M1)
eg dydx=dydu×dudx, u=x3+x, u′=3x2+1
dydx=32(x3+x)12(3x2+1)(=32√x3+x(3x2+1)) A2 N3
[3 marks]
integrating by inspection from (a) or by substitution (M1)
eg 23∫32(3x2+1)√x3+xdx, u=x3+x, dudx=3x2+1, ∫u12, u321.5
correct integrated expression in terms of x A2 N3
eg 23(x3+x)32+C, (x3+x)1.51.5+C
[3 marks]
integrating and subtracting functions (in any order) (M1)
eg ∫g−f, ∫f−∫g
correct integral (including limits, accept absence of dx) A1 N2
eg ∫10(g−f)dx, ∫106−3x2√x3+x−√x3+xdx, ∫10g(x)−∫10f(x)
[2 marks]
recognizing √x3+x is a common factor (seen anywhere, may be seen in part (c)) (M1)
eg (−3x2−1)√x3+x, ∫6−(3x2+1)√x3+x, (3x2−1)√x3+x
correct integration (A1)(A1)
eg 6x−23(x3+x)32
Note: Award A1 for 6x and award A1 for −23(x3+x)32.
substituting limits into their integrated function and subtracting (in any order) (M1)
eg 6−23(13+1)32, 0−[6−23(13+1)32]
correct working (A1)
eg 6−23×2√2, 6−23×√4×√2
area of R=6−4√23(=6−23√8,6−23×232,18−4√23) A1 N3
[6 marks]