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How do you reverse differentiation to find areas, volumes and total change?

Indefinite and definite integrals, areas under curves, integrals of standard functions, integration by substitution and by parts, integration using partial fractions, and differential equations.

A focused answer to the Edexcel A-Level Mathematics integration content, covering indefinite and definite integrals, areas under curves, standard integrals, integration by substitution and by parts, integration with partial fractions, and solving differential equations.

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What this dot point is asking

Edexcel wants you to integrate standard functions, find indefinite and definite integrals, calculate areas under and between curves, integrate by substitution and by parts, use partial fractions to integrate rational functions, and solve first-order differential equations by separating variables.

The answer

Indefinite and definite integrals

Standard integrals include exdx=ex+c\int e^x\,dx = e^x + c, 1xdx=lnx+c\int \tfrac{1}{x}\,dx = \ln|x| + c, cosxdx=sinx+c\int \cos x\,dx = \sin x + c and sinxdx=cosx+c\int \sin x\,dx = -\cos x + c.

Substitution and by parts

Substitution reverses the chain rule: replace part of the integrand with uu and convert dxdx using dudx\dfrac{du}{dx}. The aim is to choose uu so that the integral becomes a standard one in uu. For a definite integral you also change the limits into uu-values, which saves converting back to xx at the end.

Integration using partial fractions

A rational function whose denominator factorises can be split into simpler fractions, each of which integrates to a logarithm. This is the standard route for integrands such as 1(xa)(xb)\dfrac{1}{(x - a)(x - b)}.

Areas

The area between a curve and the xx-axis is abydx\int_a^b y\,dx. Where the curve dips below the axis the integral is negative, so split the region at the roots and add the magnitudes. The area between two curves y=f(x)y = f(x) and y=g(x)y = g(x) with fgf \ge g is ab(f(x)g(x))dx\int_a^b (f(x) - g(x))\,dx.

Differential equations

A separable equation dydx=f(x)g(y)\dfrac{dy}{dx} = f(x)g(y) is solved by writing 1g(y)dy=f(x)dx\int \dfrac{1}{g(y)}\,dy = \int f(x)\,dx and integrating each side, then using a boundary condition to find the constant. Such equations model situations where the rate of change depends on the current value, such as cooling, population growth and the discharge of a capacitor.

Examples in context

Try this

Q1. Find (6x24x+1)dx\int (6x^2 - 4x + 1)\,dx. [2 marks]

  • Cue. 2x32x2+x+c2x^3 - 2x^2 + x + c.

Q2. Evaluate 121xdx\int_1^2 \dfrac{1}{x}\,dx. [3 marks]

  • Cue. [lnx]12=ln2ln1=ln2[\ln x]_1^2 = \ln 2 - \ln 1 = \ln 2.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20186 marksThe region RR is bounded by the curve y=x24x+3y = x^2 - 4x + 3 and the xx-axis. Find the area of RR.
Show worked answer →

Find where the curve meets the xx-axis: x24x+3=0x^2 - 4x + 3 = 0, so (x1)(x3)=0(x - 1)(x - 3) = 0, giving x=1x = 1 and x=3x = 3 (M1, A1).

Between x=1x = 1 and x=3x = 3 the curve is below the axis, so the integral is negative and the area is its magnitude. Integrate (M1): 13(x24x+3)dx=[x332x2+3x]13\int_1^3 (x^2 - 4x + 3)\,dx = \left[\dfrac{x^3}{3} - 2x^2 + 3x\right]_1^3 (A1).

At x=3x = 3: 918+9=09 - 18 + 9 = 0. At x=1x = 1: 132+3=43\dfrac{1}{3} - 2 + 3 = \dfrac{4}{3}. The integral is 043=430 - \dfrac{4}{3} = -\dfrac{4}{3} (M1).

The area is 43=43\left|-\dfrac{4}{3}\right| = \dfrac{4}{3} square units (A1).

Markers reward the roots, integrating correctly, evaluating at the limits, and taking the magnitude for area.

Edexcel 20215 marksUse integration by parts to find xcosxdx\int x\cos x \,dx.
Show worked answer →

Choose u=xu = x (simplifies on differentiating) and dvdx=cosx\dfrac{dv}{dx} = \cos x (M1), so dudx=1\dfrac{du}{dx} = 1 and v=sinxv = \sin x (A1).

Apply the formula udvdxdx=uvvdudxdx\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx (M1):
xcosxdx=xsinxsinxdx\int x\cos x\,dx = x\sin x - \int \sin x \,dx (A1).

Integrate the remaining term: xcosxdx=xsinx+cosx+c\int x\cos x\,dx = x\sin x + \cos x + c (A1).

Markers reward a sensible choice of uu and vv, applying the by-parts formula, and the constant of integration.

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