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1.8 Integration

Syllabus
9709–2028–2029
Topic
1.8
Level
AS

Learning objectives

1.8.1Basic integration• understand integration as the reverse process of differentiation, and integrate (ax + b)n (for any rational n except -1), together with constant multiples, sums and differences e.g. xx x25 1 d3 - +;^h, x x23 1 d2+> ^h.1.8.2Constant of integration• solve problems involving the evaluation of a constant of integration e.g. to find the equation of the curve through (1, -2) for which d d 21.x y x=+1.8.3Definite integrals• evaluate definite integrals Including simple cases of 'improper' integrals, such as dxx 0 1 2 1- $ and 3 dxx2- 1;.1.8.4Area and volume• use definite integration to find - the area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line or between two curves - a volume of revolution about one of the axes. A volume of revolution may involve a region not bounded by the axis of rotation, e.g. the region between y = 9 - x2 and y = 5 rotated about the x-axis.

Reverse the power and chain rules when integrating

For rational $n e-1$:\int x^n,dx= rac{x^{n+1}}{n+1}+C,\qquad \int(ax+b)^n,dx= rac{(ax+b)^{n+1}}{a(n+1)}+C\quad(a
e0).

Rewrite roots and reciprocals as rational powers, split constant multiples, sums and differences, then integrate term by term. For a linear inner expression, divide by its gradient aa.

\int 3(2x-1)^{1/2},dx=(2x-1)^{3/2}+C,because differentiating the result gives $3(2x-1)^{1/2}$.

Differentiate the antiderivative to check every coefficient and power. Include +C+C for an indefinite integral because all constants have derivative zero.

The rule excludes n=−1n=-1; ∫x−1dx\int x^{-1}dx is logarithmic content introduced later. Also keep the real domain of fractional powers in view.

The constant of integration is fixed by a condition or a known point

After integrating a derivative, +C represents the unknown vertical position. A condition such as y=4 when x=1 determines C.

Integrate first, then substitute the given coordinate or initial value. In a motion problem, use the condition on displacement or velocity at the stated time.

If dy/dx=6x−2 and y(1)=5, then y=3x²−2x+C, giving C=4 and y=3x²−2x+4.

Setting C=0 assumes a particular origin that the question may not give; it is not a harmless simplification.

Evaluate definite integrals and replace an improper endpoint by a limit

If $F'(x)=f(x)$ and the integrand is defined on $[a,b]$,\int_a^b f(x),dx=F(b)-F(a).Constantsofintegrationcancel.Constants of integration cancel.

Find an antiderivative, substitute the upper limit and subtract the value at the lower limit. Reverse limits reverse the sign; equal limits give zero.

When $x^{-1/2}$ is undefined at $0$, approach from inside the interval:\int_0^1x^{-1/2},dx=\lim_{\varepsilon o0^+}[2x^{1/2}]{\varepsilon}^{1}=\lim{\varepsilon o0^+}(2-2\sqrt\varepsilon)=2.

A simple improper integral converges only if the one-sided limiting value is finite. Never substitute an endpoint where the integrand or antiderivative expression is undefined.

A definite integral is signed accumulation, not automatically total geometric area. If total area is requested, split at crossings and make each piece positive in the next objective.

Integrate the geometric height or cross-sectional area

Quantity Integrand for vertical slices
Area between curves upper −- lower
Volume about the xx-axis, region touches axis πy2\pi y^2 (disc)
Volume about the xx-axis, region away from axis π(R2−r2)\pi(R^2-r^2) (washer)

Sketch or compare the boundaries, solve intersections to obtain limits, identify which curve is upper/outer on each interval, split wherever that order or sign changes, then integrate and state square or cubic units.

The region between $y=9-x^2$ and $y=5$ for $-2\le x\le2$, rotated about the $x$-axis, givesV=\pi\int_{-2}^{2}[(9-x^2)^2-5^2],dx.Theinnerradiusisnon−zerobecausetheregiondoesnottouchtheaxis.The inner radius is non-zero because the region does not touch the axis.

For rotation about the yy-axis, express the horizontal radius in terms of yy and integrate the corresponding disc/washer cross-sectional area with respect to yy when the syllabus problem requires it.

Area uses a difference of heights; a washer volume uses a difference of squared radii. Do not square the difference. Shell methods are not required for this Paper 1 objective.