Challenging +1.2 This question combines improper integrals with substitution, requiring students to handle the given substitution u=√x (finding du/dx and changing limits), deal with the infinity limit properly, and integrate a rational function. While it involves multiple techniques and careful limit handling, the substitution is provided and the resulting integral is straightforward, making it moderately above average difficulty but not requiring novel insight.
State \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\) or \(du = \frac{1}{2\sqrt{x}}\,dx\)
B1
Substitute throughout for \(x\) and \(dx\)
M1
Obtain a correct integral with integrand \(\frac{2}{u^2+1}\)
A1
Integrate and obtain term of the form \(k\tan^{-1}u\)
M1
\(\left(2\tan^{-1}u\right)\)
Use limits \(\sqrt{3}\) and \(\infty\) for \(u\) or equivalent and evaluate trig
A1
e.g. \(2\left(\frac{\pi}{2} - \frac{\pi}{3}\right)\) Must be working in radians
Obtain answer \(\frac{1}{3}\pi\)
A1
Or equivalent single term
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| State $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ or $du = \frac{1}{2\sqrt{x}}\,dx$ | B1 | |
| Substitute throughout for $x$ and $dx$ | M1 | |
| Obtain a correct integral with integrand $\frac{2}{u^2+1}$ | A1 | |
| Integrate and obtain term of the form $k\tan^{-1}u$ | M1 | $\left(2\tan^{-1}u\right)$ |
| Use limits $\sqrt{3}$ and $\infty$ for $u$ or equivalent and evaluate trig | A1 | e.g. $2\left(\frac{\pi}{2} - \frac{\pi}{3}\right)$ Must be working in radians |
| Obtain answer $\frac{1}{3}\pi$ | A1 | Or equivalent single term |