Easy -2.5 This is a trick question with identical integration limits (1/2 to 1/2), making the integral trivially zero regardless of the integrand. While completing the square and recognizing the arctan form would normally be standard FP2 fare (difficulty ~0.0), spotting that no calculation is needed makes this significantly easier than average—it tests careful reading rather than integration technique.
For substituting limits in any \(\tan^{-1}\) expression
A1
For correct value
[5]
$= \int_1^{3/2} \frac{1}{(2x-1)^2+4} dx$ OR $\frac{1}{4}\int_1^{3/2} \frac{1}{(x-\frac{1}{2})^2+1} dx$ | B1 | For correct denominator (in 2nd case must include $\frac{1}{4}$)
| M1 | For integration to $k\tan^{-1}(ax+b)$ or $k\ln\left|\frac{ax+b-c}{ax+b+c}\right|$
$= \frac{1}{2}\left[\frac{1}{2}\tan^{-1}\frac{2x-1}{2}\right]_1^{3/2}$ OR $\frac{1}{4}\left[\tan^{-1}\left(x-\frac{1}{2}\right)\right]_1^{3/2}$ | A1 | FT for $ax+b$ from their denominator; For correct integration
$= \frac{1}{4}\left(\tan^{-1}1 - \tan^{-1}0\right) = \frac{1}{16}\pi$ | M1 | For substituting limits in any $\tan^{-1}$ expression
| A1 | For correct value
| [5] |
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By first completing the square in the denominator, find the exact value of
$$\int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{4x^2 - 4x + 5} dx.$$ [5]
\hfill \mbox{\textit{OCR FP2 2012 Q2 [5]}}