Integration

📚The mathematics

The basic rule

Add one to the power, then divide by the new power — the exact reverse of differentiating. The exclusion \( n \ne -1 \) exists because that case would divide by zero; \( \displaystyle\int \dfrac{1}{x}\,\mathrm{d}x \) is not on Paper 1 and waits until P2/3 5.

As with differentiation, rewrite before you integrate: \( \sqrt{x} = x^{1/2} \), \( \dfrac{3}{x^{2}} = 3x^{-2} \), and divide through by a single-term denominator.

Reversing the chain rule

Integrate as usual, then divide by the coefficient of \(x\). So \( \displaystyle\int (3x+1)^{4}\,\mathrm{d}x = \dfrac{(3x+1)^{5}}{15} + c \). This works only when the bracket is linear; for anything else you need the substitution methods of P2/3.

Definite integrals

No constant of integration is needed, because it cancels. Set the working out with square brackets and the limits, then substitute the top limit minus the bottom — in that order. When the lower limit is negative, bracket the whole substitution: forgetting to distribute the minus sign across it is the most common arithmetic error in this topic.

Areas

The area between the curve and the \(x\)-axis from \(a\) to \(b\) is \( \displaystyle\int_a^b y\,\mathrm{d}x \), provided the curve stays above the axis.

Where the curve dips below, the integral returns a negative value for that portion, and if you integrate straight through, the parts cancel and the answer is too small. For an area, find where the curve crosses the axis, split the integral at each crossing, and add the magnitudes.

For the area between two curves:

where \(a\) and \(b\) are the \(x\)-coordinates of the intersections. Subtracting first and integrating once is both quicker and safer than integrating separately and subtracting, and it handles regions below the axis automatically — provided you have identified which curve is on top. Test a single \(x\)-value between the limits to be sure.

Improper integrals

A definite integral is improper when one of its limits is infinite, or when the integrand is undefined at one of the limits. Simple cases of both appear in 9709, and they are handled the same way: integrate as usual, then consider what happens as the awkward limit is approached.

With an infinite upper limit, ask what the antiderivative tends to:

because \( \dfrac{1}{x} \to 0 \) as \( x \to \infty \). The area stretches out forever and still totals 1 — the curve falls away fast enough for the tail to contribute almost nothing.

With an integrand undefined at the lower limit, the same idea applies at the other end:

Here \( x^{-1/2} \) shoots off to infinity as \( x \to 0 \), yet the antiderivative \( 2\sqrt{x} \) is perfectly well behaved there, so the area is finite. An infinitely tall region can have a finite area, just as an infinitely long one can.

Not every improper integral has a value. For \( \displaystyle\int_1^{\infty} \frac{1}{x}\,\mathrm{d}x \) the antiderivative is \( \ln x \), and \( \ln x \to \infty \), so the integral does not converge — there is no answer to give. Saying so, with the reason, is the answer.

Volumes of revolution

Three things to get right. It is \( y^{2} \), so square first, then integrate — and squaring a bracket means expanding it, not squaring each term. For rotation about the \(y\)-axis you must rearrange to get \( x^{2} \) in terms of \(y\), and the limits must be \(y\)-values. And do not lose the \( \pi \); carrying it outside the integral throughout is the tidiest way to keep it.

For a volume between two curves, subtract the volumes: \( \pi\displaystyle\int \left(y_1^{2} - y_2^{2}\right)\mathrm{d}x \). Note that is the difference of the squares, not the square of the difference.