Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 17

For $b>0$, split the interval into $n$ equal widths $\Delta x=b/n$. Since $x^2$ increases on this interval, the left and right rectangles give lower and upper Riemann sums:Using the finite sum of squares,The rectangle bounds differ by $U_n-L_n=b^3/n\to 0$. The squeeze theorem therefore gives, directly from elementary areas,For $b=0$ the result is zero. If negative $b$ is allowed, reverse the integration limits and use the reflection $x=-u$; the same signed formula holds. No antiderivative is needed for this first-principles calculation.

Set $x=2\sin \theta$ with $-\pi /2<\theta <\pi /2$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$. The integration by substitution givesThis uses the principal inverse sine; the chosen angle interval keeps the square root positive.

Complete the square in the denominator and split the numerator into a multiple of its derivative plus a constant:ThusThe natural logarithm needs no absolute-value correction because its quadratic argument is strictly positive for every real $x$. The second term is an inverse tangent antiderivative.

First apply integration by parts with $u=x^k$ and $dv=\sin xdx$:For $k>1$, integrate the remaining integral by parts once more. Its boundary term $[x^{k-1}\sin x{]}_0^{\pi }$ is zero, soAt $k=1$, direct integration gives $F(1)=\pi$, agreeing with the same formula; the coefficient of $F(-1)$ vanishes and that integral is finite. Now follow the odd chain:This is the sine-weighted power integral recurrence.