Challenging +1.2 This is a standard improper integral requiring integration by parts and handling the logarithmic singularity at x=0. While it requires knowledge of the limit of x ln x as xβ0+ and careful presentation of the limiting process, it follows a well-established technique taught in Further Maths. The 6 marks reflect the need to show all steps clearly, but the problem itself is a textbook example with no novel insight requiredβmoderately above average difficulty due to the improper integral aspect and integration by parts, but still routine for Further Maths students.
Question 10:
10 | Defines the improper integral
as a limit | 2.4 | E1 | 8 8
οΏ½ln(π₯π₯)dπ₯π₯ = βliβm0οΏ½ln(π₯π₯)dπ₯π₯
0 β
β²
π’π’ = lnπ₯π₯ π£π£ = 1
β² 1
π’π’ = π₯π₯ π£π£ = π₯π₯
8
8
8
οΏ½ln(π₯π₯)dπ₯π₯ = βliβm0οΏ½[π₯π₯ln(π₯π₯)]β βοΏ½ 1dπ₯π₯οΏ½
β
0
8
= βliβm0([π₯π₯ln(π₯π₯)βπ₯π₯]β)
= {8ln(8)β8}ββliβm0{β ln (β) ββ}
As
βliβm0{β ln (β) } = 0
Then
8
β«0 ln(π₯π₯)dπ₯π₯ = 8ln(8)β8
Selects and uses the method
of integration by parts.
Implied by stating and using
the formula for the integral of | 3.1a | M1
l nπ₯π₯
Obtains the correct integral
with or without c.
Condone no limits | 1.1b | A1
Substitutes 8 correctly into
their two-term expression for
the integral | 1.1a | M1
Applies the limiting process
correctly, using
This does not have to be
βliβm0{β ln (β) } = 0
stated explicitly | 2.2a | M1
Obtains correct value OE,
explicitly stating
βliβm0{β ln (β) } = 0
NMS = 0 | 1.1b | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical solution
Evaluate the improper integral
$$\int_0^8 \ln x \, \mathrm{d}x$$
showing the limiting process.
[6 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q10 [6]}}