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A LEVEL (P3) INTEGRATION MARK SCHEME 2023 UPDATE
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A LEVEL (P3) INTEGRATION MARK SCHEME 2023 UPDATE
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A LEVEL (P3)INTEGRATION
MARK SCHEME
, MARKING SCHEMES 1 TOPIC 8: INTEGRATION
2 (i) Use product or quotient rule M1*
1 1
− x 1 − x
Obtain first derivative 2 xe 2 − x 2e 2 or equivalent A1
2
Equate derivative to zero and solve for non-zero x M1(dep*)
Obtain answer x = 4 A1 4
1 1
− x − x
∫
2
(ii) Integrate by parts once, obtaining kx e 2 +l xe 2 dx , where kl ≠ 0 M1
1 1
− x − x
Obtain integral − 2 x 2 e 2
∫
+ 4 xe 2 dx , or any unsimplified equivalent A1
1
Complete the integration, obtaining − 2 x + 4 x + 8 ( 2
or equivalent )
e
− x
2 A1
Having integrated by parts twice, use limits x = 0 and x = 1 in the complete integral M1
1
−
Obtain simplified answer 16 − 26e 2 or equivalent A1 5
A Bx + C
3 (a)(i) State answer + 2 B1 1
x +4 x +3
A Bx + C A B C
(ii) State answer + or + + B2 2
x − 2 (x + 2)2 x − 2 x + 2 (x + 2)2
A B Cx + D
[Award B1 if the B term is omitted or for the form + + .]
x − 2 x + 2 (x + 2)2
A B
(b) Stating or implying f(x) ≡ + , use a relevant method to determine A or B M1
x +1 x − 2
Obtain A = 1 and B = 2 A1
[SR: If A = 1 and B = 2 stated without working, award B1 + B1.]
Integrate and obtain terms ln (x + 1) + 2 ln (x − 2) A1√ + A1√
Use correct limits correctly in the complete integral M1
Obtain given answer ln 5 following full and exact working A1 6
, MARKING SCHEMES 2 TOPIC 8: INTEGRATION
dx
4 (i) State or imply dx = sec 2θ dθ or = sec 2θ B1
dθ
Substitute for x and dx throughout the integral M1
Obtain integral in terms of θ in any correct form A1
Reduce to the given form correctly A1 4
1
(ii) State integral 2
sin 2θ B1
Use limits θ = 0 and θ = 1
4
π correctly in integral of the form k sin 2θ M1
Obtain answer ½ or 0.5 A1 3
5 (i) Use quotient or product rule M1
Obtain derivative in any correct form A1
Equate derivative to zero and solve for x or x2 M1
Obtain x = 1 correctly A1 4
[Differentiating (x 2 + 1)y = x using the product rule can also earn the
first M1A1.]
[SR: if the quotient rule is misused, with a ‘reversed’ numerator or v
instead of v² in the denominator, award M0A0 but allow the
following M1A1.]
(ii) Obtain indefinite integral of the form k ln ( x 2 + 1) , where k = ½, 1 or 2 M1*
Use limits x = 0 and x = p correctly, or equivalent M1(dep*)
Obtain answer ½ ln(p2 +1) A1 3
[Also accept –ln cos θ or ln cos θ , where x = tan θ , for the first M1*.]
(iii) Equate to 1 and convert equation to the form p2 + 1 = exp(1/k ) M1
Obtain answer p = 2.53 A1 2
dx
6 (i) State = 2sin θ cos θ , or dx = 2sin θ cos θ dθ B1
dθ
Substitute for x and dx throughout M1
Obtain any correct form in terms of θ A1
Reduce to the given form correctly A1 [4]
(ii) Use cos 2A formula, replacing integrand by a + bcos 2θ , where ab ≠ 0 M1*
Integrate and obtain θ − 12 sin 2θ A1√
Use limits θ = 0 and θ = 16 π M1(dep*)
Obtain exact answer 1π − 14 3 , or equivalent A1 [4]
6
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