Edexcel A-level Maths coverage

Integration

Section 8
8 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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8.1

Know and use the Fundamental Theorem of Calculus.

  • Integration is the reverse process of differentiation: if dydx=f(x)\frac{\mathrm dy}{\mathrm dx}=f(x), then y=f(x)dxy=\int f(x)\,\mathrm dx recovers the family of curves with that gradient function.
  • Every indefinite integral needs a constant of integration: f(x)dx=F(x)+c\int f(x)\,\mathrm dx=F(x)+c. One known point on the curve fixes the value of cc.
  • A definite integral is evaluated from any antiderivative: abf(x)dx=F(b)F(a)\int_a^b f(x)\,\mathrm dx=F(b)-F(a); the constant of integration cancels in the subtraction.
  • A common error is to omit +c+c from an indefinite integral, or to substitute the known point into f(x)f(x) instead of the integrated function when finding cc.

Tier 1 · Easy

3 marks
ORIGINAL

A curve has gradient function dydx=6x24\frac{\mathrm dy}{\mathrm dx}=6x^2-4 and passes through the point (1,5)(1,5). Find the equation of the curve.

Tier 2 · Standard

4 marks
ORIGINAL

The function ff satisfies f(x)=3x2x2f'(x)=3\sqrt{x}-\dfrac{2}{x^2} for x>0x>0, and f(4)=10f(4)=10. Find f(x)f(x).

Tier 3 · Hard

5 marks
ORIGINAL

A curve y=f(x)y=f(x) satisfies f(x)=6x+2f''(x)=6x+2. The gradient of the curve at x=1x=1 is 33, and the curve passes through (2,4)(2,4). Find f(x)f(x), and verify your answer by evaluating 12f(x)dx\int_1^2 f'(x)\,\mathrm dx and comparing it with f(2)f(1)f(2)-f(1).

8.2

Integrate xⁿ (excluding n = −1) and related sums, differences and constant multiples; integrate e^(kx), 1/x, sin kx, cos kx and related sums, differences and constant multiples.

  • For n1n\ne-1, xndx=xn+1n+1+C\int x^n\,\mathrm dx=\frac{x^{n+1}}{n+1}+C; rewrite roots and reciprocals as powers before applying the rule.
  • Standard forms include ekxdx=1kekx+C\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C and 1xdx=lnx+C\int\frac1x\,\mathrm dx=\ln|x|+C on intervals not crossing 00.
  • Divide by the inner coefficient in trigonometric integrals. Also use sec2(kx)dx=1ktan(kx)+C\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C, tanxdx=lncosx+C\int\tan x\,\mathrm dx=-\ln|\cos x|+C, and identities for powers such as sin2x=12(1cos2x)\sin^2x=\frac12(1-\cos2x).
  • A common error is to forget the constant of integration or to multiply by kk rather than divide by it when integrating a function of kxkx.

Tier 1 · Easy

4 marks
ORIGINAL

Find (6x24x1/2+3sec2(2x))dx\int(6x^2-4x^{1/2}+3\sec^2(2x))\,\mathrm dx.

Tier 2 · Standard

6 marks
ORIGINAL

Find (6e3x+4x5sin(2x)+3tanx)dx\int\left(6e^{3x}+\frac4x-5\sin(2x)+3\tan x\right)\,\mathrm dx.

Tier 3 · Hard

6 marks
ORIGINAL

Use trigonometric identities to find (4sin2x+3tan2x)dx\int(4\sin^2x+3\tan^2x)\,\mathrm dx.

8.3

Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

  • Evaluate abf(x)dx\int_a^b f(x)\,\mathrm dx by finding an antiderivative FF and calculating F(b)F(a)F(b)-F(a).
  • A definite integral is signed area. Split at roots or intersections if the required geometric area includes regions below the axis or where the upper curve changes.
  • For area between curves, solve their intersection equations first and integrate yupperylowery_{\text{upper}}-y_{\text{lower}} over each relevant interval.
  • A common error is to reverse the subtraction or to quote a negative definite integral as a negative geometric area.

Tier 1 · Easy

3 marks
ORIGINAL

Find the area between the line y=2x+1y=2x+1, the xx-axis, and the lines x=0x=0 and x=3x=3.

Tier 2 · Standard

5 marks
ORIGINAL

Find the finite area enclosed by the curve y=4xx2y=4x-x^2 and the line y=xy=x.

Tier 3 · Hard

6 marks
ORIGINAL

A curve is given parametrically by x=t2x=t^2 and y=t3y=t^3 for 0t20\leq t\leq2. Find the exact area between the curve, the xx-axis and the line x=4x=4.

8.4

Understand and use integration as the limit of a sum.

  • Partition an interval into strips of width Δx\Delta x and form f(xr)Δx\sum f(x_r)\Delta x; its limit as the maximum strip width tends to zero is the definite integral.
  • To recognise a limit, identify the factor playing the role of Δx\Delta x and rewrite the sampled expression in terms of an endpoint such as xr=a+rΔxx_r=a+r\Delta x.
  • A sum with factor 1/n1/n usually samples an interval of length 11; a different interval width introduces a corresponding scale factor.
  • A common error is to identify the integrand but omit the width factor, which changes the value of the limiting integral.

Tier 1 · Easy

4 marks
ORIGINAL

Express 02x2dx\int_0^2x^2\,\mathrm dx as a limit of a right-endpoint sum and evaluate it.

Tier 2 · Standard

5 marks
ORIGINAL

Evaluate limn1nr=1n(1+3rn)2\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2 by expressing it as a definite integral.

Tier 3 · Hard

6 marks
ORIGINAL

Evaluate exactly limn2nr=1nln(1+2rn)\lim_{n\to\infty}\frac2n\sum_{r=1}^n\ln\left(1+\frac{2r}{n}\right).

8.5

Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively.

  • Substitution reverses the chain rule: choose u=g(x)u=g(x) so the remaining factor supplies du=g(x)dx\mathrm du=g'(x)\,\mathrm dx. In particular, recognise f(x)f(x)dx=lnf(x)+C\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C.
  • Integration by parts reverses the product rule: udv=uvvdu\int u\,\mathrm dv=uv-\int v\,\mathrm du; choose uu so that differentiating it simplifies the integral.
  • For a definite integral, either change the limits to the new variable or return fully to the original variable before applying the old limits.
  • A common error is to omit the minus sign in integration by parts or to mix xx and uu in the same transformed integral.

Tier 1 · Easy

3 marks
ORIGINAL

Find 6x3x2+5dx\int\frac{6x}{3x^2+5}\,\mathrm dx.

Tier 2 · Standard

5 marks
ORIGINAL

Find xe2xdx\int xe^{2x}\,\mathrm dx.

Tier 3 · Hard

7 marks
ORIGINAL

Evaluate exactly 01xln(1+x2)dx\int_0^1x\ln(1+x^2)\,\mathrm dx.

8.6

Integrate using partial fractions that are linear in the denominator.

  • Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants.
  • For Axa\frac{A}{x-a}, the integral is Alnxa+CA\ln|x-a|+C; retain absolute values unless the domain makes the sign known.
  • Determine constants by equating coefficients or substituting convenient values that make all but one term vanish.
  • A common error is to integrate the unfactorised denominator as though f(x)/f(x)dx\int f'(x)/f(x)\,\mathrm dx applied when the numerator is not its derivative.

Tier 1 · Easy

3 marks
ORIGINAL

Find (3x1+2x+2)dx\int\left(\frac3{x-1}+\frac2{x+2}\right)\,\mathrm dx.

Tier 2 · Standard

5 marks
ORIGINAL

Express 5x+1(x1)(x+2)\frac{5x+1}{(x-1)(x+2)} in partial fractions and hence integrate it.

Tier 3 · Hard

7 marks
ORIGINAL

Evaluate exactly 015x+1(x+1)(x+2)dx\int_0^1\frac{5x+1}{(x+1)(x+2)}\,\mathrm dx.

8.7

Evaluate the analytical solution of simple first order differential equations with separable variables, including finding particular solutions.

  • For dydx=g(x)h(y)\frac{\mathrm dy}{\mathrm dx}=g(x)h(y), rearrange to place all yy-terms with dy\mathrm dy and all xx-terms with dx\mathrm dx, then integrate both sides.
  • Include a constant after integration and use the initial condition to determine it. A family of solutions can be sketched from its equilibrium curves, initial values, gradients and long-term behaviour.
  • When logarithms arise, exponentiate carefully and use the stated domain or initial condition to choose any required sign or branch.
  • A common error is to divide by a factor involving yy without checking whether doing so loses a constant equilibrium solution.

Tier 1 · Easy

4 marks
ORIGINAL

Solve dydx=2xy\frac{\mathrm dy}{\mathrm dx}=2xy, given that y=3y=3 when x=0x=0.

Tier 2 · Standard

5 marks
ORIGINAL

Solve dydx=x+1y\frac{\mathrm dy}{\mathrm dx}=\frac{x+1}{y}, given that y=2y=2 when x=0x=0 and y>0y>0.

Tier 3 · Hard

9 marks
ORIGINAL

Solve dydx=y(4y)\frac{\mathrm dy}{\mathrm dx}=y(4-y), given that y=1y=1 when x=0x=0. State the equilibrium solutions and describe the particular solution's long-term behaviour.

8.8

Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics.

  • Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.
  • In kinematics, v=dsdtv=\frac{\mathrm ds}{\mathrm dt} and a=dvdta=\frac{\mathrm dv}{\mathrm dt}; the sign of velocity gives direction, while a change of sign marks a reversal of motion.
  • Check whether the mathematical solution remains meaningful on the time interval and within the physical range assumed by the model.
  • A common error is to state only that a model is 'unrealistic'; name a specific assumption, such as constant environmental conditions or neglect of resistance, and explain its effect.

Tier 1 · Easy

4 marks
ORIGINAL

A population model has solution N=500e0.12tN=500e^{0.12t}, where tt is measured in years. Interpret the constants 500500 and 0.120.12, and state one limitation of the model.

Tier 2 · Standard

6 marks
ORIGINAL

The velocity of a particle is modelled by v=20(1e0.5t)m s1v=20(1-e^{-0.5t})\,\text{m s}^{-1} for t0t\geq0. Find the limiting velocity and the time when v=15m s1v=15\,\text{m s}^{-1}. State one limitation of the model.

Tier 3 · Hard

8 marks
ORIGINAL

A particle has velocity v=1824e0.3tm s1v=18-24e^{-0.3t}\,\text{m s}^{-1} for t0t\geq0, with displacement s=0s=0 at t=0t=0. Find when the particle changes direction, its displacement then, and its limiting velocity. Give one limitation of using this model for arbitrarily large tt.