| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Improper integral to infinity with inverse trig |
| Difficulty | Challenging +1.2 This is a straightforward Further Maths improper integral question requiring recognition of the standard inverse tan form, completing the integration with correct limits including infinity, and simple arithmetic to find k. While it involves Further Maths content (improper integrals and inverse trig integration), the execution is mechanical with no novel problem-solving required. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Because the upper limit is infinite | B1 | Suitable explanation that one bound/limit is infinite, e.g. "one of the limits is unbounded"; do not allow if they only say limit/integral is undefined without further explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int\frac{1}{9x^2+16}\,dx=\frac{1}{12}\arctan\!\left(\frac{3x}{4}\right)\) | M1, A1 | Integrates to \(\alpha\arctan(\beta x)\) where \(\beta\neq 1\); correct integration unsimplified or simplified |
| \(\int_{\frac{4}{3}}^{\infty}\frac{1}{9x^2+16}\,dx=\frac{1}{12}\lim_{t\to\infty}\left[\arctan\!\left(\frac{3x}{4}\right)\right]_{\frac{4}{3}}^{t}\) | dM1 | Applies correct limits \(t\) and \(\frac{4}{3}\) with evidence of applying infinite limit to obtain non-zero value; dependent on first M |
| \(=\frac{1}{12}\!\left(\lim_{t\to\infty}\arctan\!\left(\frac{3t}{4}\right)-\arctan(1)\right)=\frac{1}{12}\!\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{48}\) | A1 | Correct value with evidence of limiting process on upper bound; must see \(\lim_{t\to\infty}\) at some stage |
# Question 3(a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Because the upper limit is infinite | B1 | Suitable explanation that one bound/limit is infinite, e.g. "one of the limits is unbounded"; do not allow if they only say limit/integral is undefined without further explanation |
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# Question 3(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\int\frac{1}{9x^2+16}\,dx=\frac{1}{12}\arctan\!\left(\frac{3x}{4}\right)$ | M1, A1 | Integrates to $\alpha\arctan(\beta x)$ where $\beta\neq 1$; correct integration unsimplified or simplified |
| $\int_{\frac{4}{3}}^{\infty}\frac{1}{9x^2+16}\,dx=\frac{1}{12}\lim_{t\to\infty}\left[\arctan\!\left(\frac{3x}{4}\right)\right]_{\frac{4}{3}}^{t}$ | dM1 | Applies correct limits $t$ and $\frac{4}{3}$ with evidence of applying infinite limit to obtain non-zero value; dependent on first M |
| $=\frac{1}{12}\!\left(\lim_{t\to\infty}\arctan\!\left(\frac{3t}{4}\right)-\arctan(1)\right)=\frac{1}{12}\!\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{48}$ | A1 | Correct value with evidence of limiting process on upper bound; must see $\lim_{t\to\infty}$ at some stage |\begin{enumerate}
\item (a) Explain why
\end{enumerate}
$$\int _ { \frac { 4 } { 3 } } ^ { \infty } \frac { 1 } { 9 x ^ { 2 } + 16 } d x$$
is an improper integral.\\
(b) Show that
$$\int _ { \frac { 4 } { 3 } } ^ { \infty } \frac { 1 } { 9 x ^ { 2 } + 16 } d x = k \pi$$
where $k$ is a constant to be determined.
\hfill \mbox{\textit{Edexcel CP2 2024 Q3 [5]}}