Challenging +1.8 This is a Further Maths integration question requiring completion of the square, hyperbolic substitution (likely x+3 = 2cosh(u)), and careful handling of limits including absolute values. While the techniques are standard for Further Maths, the 7-mark allocation reflects multiple steps: completing the square, choosing and executing the substitution, managing the square root of cosh²u, integrating to get sinh terms, and evaluating limits with inverse hyperbolic functions. The algebraic manipulation is substantial but follows a well-established method for this type of integral.
Question 17:
17 | Completes the square to
obtain
( x+a )2 +b | 3.1a | M1 | x2 + 6x + 8 = (x + 3)2 – 1
Let x + 3 = cosh θ
Then
d x
= sinh θ
d θ
1−
Let I = x 2 + 6 x + 8 d x
2
When x = –2, cosh θ = 1 and θ = 0
When x = 1, cosh θ = 4
and θ = cosh–1 (4)
–1 c o s h (4 )
I = s 2 in h θ θ s in h d θ
0
c o –1 s h (4 )
= s in h 2 θ d θ
0
–1 c o s h (4 )
1
I = ( c o s θ – d h 2 1 ) θ
2
0
–1 c o s h (4 )
1 θ s in h 2
= – θ
2 2
0
1 –1 c o s h (4 )
= θ s in h c o θ s h – θ
2 0
When
cosh θ = 4, sinh θ = 4 2 – 1 = 1 5
1( )
I = 4 15 –cosh–1(4) –0
2
1
=2 15 – cosh–1(4)
2
Writes
x a k + = c o s h
or
x+a=ksinh | 2.2a | M1
Obtains sinh2 θ dθ | 1.1b | A1
Uses
1 1
s in h 2 θ = c o s h 2 θ −
2 2
Condone
1 1
sinh2 θ= cosh2θ
2 2 | 3.1a | M1
1 s in h 2 θ
Obtains – θ OE
2 2 | 1.1b | A1
Uses cosh2 θ−sinh2 θ =1
when x = 1 to obtain
s in h θ = 1 5 | 2.2a | B1
Completes a reasoned
argument to obtain
1
2 15 – cosh–1(4)
2
AG | 2.1 | R1
Question total | 7
Q | Marking instructions | AO | Marks | Typical solution
By making a suitable substitution, show that
$$\int_{-2}^{1} \sqrt{x^2 + 6x + 8} \, dx = 2\sqrt{15} - \frac{1}{2}\cosh^{-1}(4)$$
[7 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q17 [7]}}