Date | May 2017 | Marks available | 3 | Reference code | 17M.1.sl.TZ1.5 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Find ∫xex2−1dx.
[4]
a.
Find f(x), given that f′(x)=xex2−1 and f(−1)=3.
[3]
b.
Markscheme
valid approach to set up integration by substitution/inspection (M1)
egu=x2−1, du=2x, ∫2xex2−1dx
correct expression (A1)
eg12∫2xex2−1dx, 12∫eudu
12ex2−1+c A2 N4
Notes: Award A1 if missing “+c”.
[4 marks]
a.
substituting x=−1 into their answer from (a) (M1)
eg12e0, 12e1−1=3
correct working (A1)
eg12+c=3, c=2.5
f(x)=12ex2−1+2.5 A1 N2
[3 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.