Find ∫√2x - 1 dx.
- (a)Find ∫√2x - 1 dx.[2]
- (b)(i)Find ∫sin 4x dx.[2]
- (b)(ii)Hence evaluate ∫(π/8)^(π/4) sin 4x dx.[2]
- (c)Show that ∫(0)^(ln8) e³ dx = 3.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [B2] (2x-1)^(3/2) / 3 [+c] oe isw (B1 for k × (2x-1)^(3/2) / (3/2) where k ≠ 0 or for (2x-1)^(3/2) / (3/2) × k where k ≠ 0)
- [M1] for k cos 4x [+c] where k < 0 or k = -1/4
- [A1] -1/4 cos 4x [+c]
- [M1] Sight of correct substitution of limits: -1/4 cos(4π/4) - (-1/4 cos(4π/8)) oe; FT their k cos 4x from (b)(i); dep on M1 awarded in (b)(i)
- [A1] 1/4
- [M1] For integration leading to k e^(x/3) [+c] (implies k e^(x/3) with k any non-zero constant)
- [A1] k = 3
- [M1] Sight of correct substitution of limits: their k e^(ln8/3) - their k e^0 oe
- [B1] Correct evaluation of e^(ln8/3) as 2
- [A1] 6 - 3 = 3
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