8 Integration — printable coverage pack

8.1 · Know and use the Fundamental Theorem of Calculus.

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

Tier 1 · Easy

1. A curve has gradient function $\frac{\mathrm dy}{\mathrm dx}=6x^2-4$ and passes through the point $(1,5)$ . Find the equation of the curve.[3 marks]
Answer
- $y=2x^3-4x+7$
Method: Integrating the gradient function gives $y=2x^3-4x+c$ . Substituting $(1,5)$ : $5=2-4+c$ , so $c=7$ . Hence $y=2x^3-4x+7$ .

Tier 2 · Standard

1. The function $f$ satisfies $f'(x)=3\sqrt{x}-\dfrac{2}{x^2}$ for $x>0$ , and $f(4)=10$ . Find $f(x)$ .[4 marks]
Answer
- $f(x)=2x^{3/2}+\dfrac{2}{x}-\dfrac{13}{2}$
Method: Rewrite as $f'(x)=3x^{1/2}-2x^{-2}$ and integrate: $f(x)=2x^{3/2}+2x^{-1}+c$ . Substituting $x=4$ : $2(8)+\frac{2}{4}+c=10$ , so $16.5+c=10$ and $c=-\frac{13}{2}$ . Hence $f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}$ .

Tier 3 · Hard

1. A curve $y=f(x)$ satisfies $f''(x)=6x+2$ . The gradient of the curve at $x=1$ is $3$ , and the curve passes through $(2,4)$ . Find $f(x)$ , and verify your answer by evaluating $\int_1^2 f'(x)\,\mathrm dx$ and comparing it with $f(2)-f(1)$ .[5 marks]
Answer
- $f(x)=x^3+x^2-2x-4$
- $\int_1^2 f'(x)\,\mathrm dx=8=f(2)-f(1)$
Method: Integrate once: $f'(x)=3x^2+2x+c_1$ ; from $f'(1)=3$ , $3+2+c_1=3$ so $c_1=-2$ . Integrate again: $f(x)=x^3+x^2-2x+c_2$ ; from $f(2)=4$ , $8+4-4+c_2=4$ so $c_2=-4$ . Then $f(2)-f(1)=4-(-4)=8$ , and $\int_1^2(3x^2+2x-2)\,\mathrm dx=\left[x^3+x^2-2x\right]_1^2=8-0=8$ , confirming $\int_a^b f'(x)\,\mathrm dx=f(b)-f(a)$ .

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 $n\ne-1$ , $\int x^n\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$ ; rewrite roots and reciprocals as powers before applying the rule.
Standard forms include $\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C$ and $\int\frac1x\,\mathrm dx=\ln|x|+C$ on intervals not crossing $0$ .
Divide by the inner coefficient in trigonometric integrals. Also use $\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C$ , $\int\tan x\,\mathrm dx=-\ln|\cos x|+C$ , and identities for powers such as $\sin^2x=\frac12(1-\cos2x)$ .
A common error is to forget the constant of integration or to multiply by $k$ rather than divide by it when integrating a function of $kx$ .

Tier 1 · Easy

1. Find $\int(6x^2-4x^{1/2}+3\sec^2(2x))\,\mathrm dx$ .[4 marks]
Answer
- $2x^3-\frac83x^{3/2}+\frac32\tan(2x)+C$
Method: The power terms give $2x^3-\frac83x^{3/2}$ . Since $\int\sec^2(2x)\,\mathrm dx=\frac12\tan(2x)$ , the final term gives $\frac32\tan(2x)$ . Add $C$ .

Tier 2 · Standard

1. Find $\int\left(6e^{3x}+\frac4x-5\sin(2x)+3\tan x\right)\,\mathrm dx$ .[6 marks]
Answer
- $2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C$
Method: Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$ . Since $\int\tan x\,\mathrm dx=-\ln|\cos x|$ , the final term gives $-3\ln|\cos x|$ . Add $C$ .

Tier 3 · Hard

1. Use trigonometric identities to find $\int(4\sin^2x+3\tan^2x)\,\mathrm dx$ .[6 marks]
Answer
- $3\tan x-x-\sin(2x)+C$
Method: Use $\sin^2x=\frac12(1-\cos2x)$ and $\tan^2x=\sec^2x-1$ . The integrand becomes $2-2\cos2x+3\sec^2x-3=3\sec^2x-1-2\cos2x$ . Integrating gives $3\tan x-x-\sin(2x)+C$ .

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

Evaluate $\int_a^b f(x)\,\mathrm dx$ by finding an antiderivative $F$ and calculating $F(b)-F(a)$ .
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 $y_{\text{upper}}-y_{\text{lower}}$ 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

1. Find the area between the line $y=2x+1$ , the $x$ -axis, and the lines $x=0$ and $x=3$ .[3 marks]
Answer
- Area $=12$ square units
Method: The line is above the axis on the interval, so the area is $\int_0^3(2x+1)\,\mathrm dx=[x^2+x]_0^3=9+3=12$ .

Tier 2 · Standard

1. Find the finite area enclosed by the curve $y=4x-x^2$ and the line $y=x$ .[5 marks]
Answer
- Area $=\frac92$ square units
Method: Intersections satisfy $4x-x^2=x$ , so $x(3-x)=0$ and $x=0,3$ . The parabola is above the line between them. Thus the area is $\int_0^3(3x-x^2)\,\mathrm dx=[\frac32x^2-\frac13x^3]_0^3=\frac{27}{2}-9=\frac92$ .

Tier 3 · Hard

1. A curve is given parametrically by $x=t^2$ and $y=t^3$ for $0\leq t\leq2$ . Find the exact area between the curve, the $x$ -axis and the line $x=4$ .[6 marks]
Answer
- Area $=\frac{64}{5}$ square units
Method: For a parametric curve, area is $\int y\,\mathrm dx=\int_0^2y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt$ . Here $\frac{\mathrm dx}{\mathrm dt}=2t$ , so the area is $\int_0^2t^3(2t)\,\mathrm dt=2\int_0^2t^4\,\mathrm dt=2[\frac15t^5]_0^2=\frac{64}{5}$ .

8.4 · Understand and use integration as the limit of a sum.

Partition an interval into strips of width $\Delta x$ and form $\sum f(x_r)\Delta x$ ; 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 $\Delta x$ and rewrite the sampled expression in terms of an endpoint such as $x_r=a+r\Delta x$ .
A sum with factor $1/n$ usually samples an interval of length $1$ ; 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

1. Express $\int_0^2x^2\,\mathrm dx$ as a limit of a right-endpoint sum and evaluate it.[4 marks]
Answer
- $\lim_{n\to\infty}\frac8{n^3}\sum_{r=1}^nr^2=\int_0^2x^2\,\mathrm dx=\frac83$
Method: Use $\Delta x=2/n$ and $x_r=2r/n$ . Then $\sum f(x_r)\Delta x=\sum_{r=1}^n(2r/n)^2(2/n)=\frac8{n^3}\sum_{r=1}^nr^2$ . Its limit is $\int_0^2x^2\,\mathrm dx=[x^3/3]_0^2=\frac83$ .

Tier 2 · Standard

1. Evaluate $\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2$ by expressing it as a definite integral.[5 marks]
Answer
- $7$
Method: With $x=r/n$ , the limit is $\int_0^1(1+3x)^2\,\mathrm dx$ . Equivalently, with $u=1+3x$ , it is $\frac13\int_1^4u^2\,\mathrm du=\frac19(4^3-1^3)=7$ .

Tier 3 · Hard

1. Evaluate exactly $\lim_{n\to\infty}\frac2n\sum_{r=1}^n\ln\left(1+\frac{2r}{n}\right)$ .[6 marks]
Answer
- $3\ln3-2$
Method: Here $\Delta x=2/n$ and the right endpoints are $x_r=2r/n$ , so the limit is $\int_0^2\ln(1+x)\,\mathrm dx$ . By substitution followed by integration by parts, an antiderivative is $(1+x)\ln(1+x)-(1+x)$ . Evaluation from $0$ to $2$ gives $(3\ln3-3)-(-1)=3\ln3-2$ .

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 $u=g(x)$ so the remaining factor supplies $\mathrm du=g'(x)\,\mathrm dx$ . In particular, recognise $\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C$ .
Integration by parts reverses the product rule: $\int u\,\mathrm dv=uv-\int v\,\mathrm du$ ; choose $u$ 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 $x$ and $u$ in the same transformed integral.

Tier 1 · Easy

1. Find $\int\frac{6x}{3x^2+5}\,\mathrm dx$ .[3 marks]
Answer
- $\ln(3x^2+5)+C$
Method: The numerator is the derivative of the denominator. Therefore this has the form $\int f'(x)/f(x)\,\mathrm dx$ , giving $\ln|3x^2+5|+C$ . Since $3x^2+5>0$ , this is $\ln(3x^2+5)+C$ .

Tier 2 · Standard

1. Find $\int xe^{2x}\,\mathrm dx$ .[5 marks]
Answer
- $e^{2x}\left(\frac x2-\frac14\right)+C$
Method: Use integration by parts with $u=x$ and $\mathrm dv=e^{2x}\,\mathrm dx$ . Then $\mathrm du=\mathrm dx$ and $v=\frac12e^{2x}$ . Hence $\int xe^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac12\int e^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac14e^{2x}+C$ .

Tier 3 · Hard

1. Evaluate exactly $\int_0^1x\ln(1+x^2)\,\mathrm dx$ .[7 marks]
Answer
- $\ln2-\frac12$
Method: Let $u=1+x^2$ , so $\mathrm du=2x\,\mathrm dx$ and the limits become $u=1$ to $u=2$ . The integral is $\frac12\int_1^2\ln u\,\mathrm du$ . By parts, $\int\ln u\,\mathrm du=u\ln u-u$ . Therefore the value is $\frac12[u\ln u-u]_1^2=\frac12(2\ln2-2+1)=\ln2-\frac12$ .

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 $\frac{A}{x-a}$ , the integral is $A\ln|x-a|+C$ ; 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 $\int f'(x)/f(x)\,\mathrm dx$ applied when the numerator is not its derivative.

Tier 1 · Easy

1. Find $\int\left(\frac3{x-1}+\frac2{x+2}\right)\,\mathrm dx$ .[3 marks]
Answer
- $3\ln|x-1|+2\ln|x+2|+C$
Method: Integrate each linear-denominator term directly: $\int3/(x-1)\,\mathrm dx=3\ln|x-1|$ and $\int2/(x+2)\,\mathrm dx=2\ln|x+2|$ . Add $C$ .

Tier 2 · Standard

1. Express $\frac{5x+1}{(x-1)(x+2)}$ in partial fractions and hence integrate it.[5 marks]
Answer
- $\frac{5x+1}{(x-1)(x+2)}=\frac2{x-1}+\frac3{x+2}$
- $2\ln|x-1|+3\ln|x+2|+C$
Method: Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$ . Then $5x+1=A(x+2)+B(x-1)$ . Substituting $x=1$ gives $A=2$ , and $x=-2$ gives $B=3$ . Integrating the decomposition gives $2\ln|x-1|+3\ln|x+2|+C$ .

Tier 3 · Hard

1. Evaluate exactly $\int_0^1\frac{5x+1}{(x+1)(x+2)}\,\mathrm dx$ .[7 marks]
Answer
- $9\ln3-13\ln2$
Method: Write $\frac{5x+1}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}$ . Then $5x+1=A(x+2)+B(x+1)$ , giving $A=-4$ and $B=9$ . The integral is $[-4\ln(x+1)+9\ln(x+2)]_0^1=(-4\ln2+9\ln3)-9\ln2=9\ln3-13\ln2$ .

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

For $\frac{\mathrm dy}{\mathrm dx}=g(x)h(y)$ , rearrange to place all $y$ -terms with $\mathrm dy$ and all $x$ -terms with $\mathrm dx$ , 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 $y$ without checking whether doing so loses a constant equilibrium solution.

Tier 1 · Easy

1. Solve $\frac{\mathrm dy}{\mathrm dx}=2xy$ , given that $y=3$ when $x=0$ .[4 marks]
Answer
- $y=3e^{x^2}$
Method: Separate variables: $\frac1y\,\mathrm dy=2x\,\mathrm dx$ . Integrating gives $\ln|y|=x^2+C$ , so $y=Ae^{x^2}$ . The condition $y(0)=3$ gives $A=3$ , hence $y=3e^{x^2}$ .

Tier 2 · Standard

1. Solve $\frac{\mathrm dy}{\mathrm dx}=\frac{x+1}{y}$ , given that $y=2$ when $x=0$ and $y>0$ .[5 marks]
Answer
- $y=\sqrt{x^2+2x+4}$
Method: Separate: $y\,\mathrm dy=(x+1)\,\mathrm dx$ . Integration gives $\frac12y^2=\frac12x^2+x+C$ , so $y^2=x^2+2x+C_1$ . The condition $y(0)=2$ gives $C_1=4$ . Since $y>0$ , $y=\sqrt{x^2+2x+4}$ .

Tier 3 · Hard

1. Solve $\frac{\mathrm dy}{\mathrm dx}=y(4-y)$ , given that $y=1$ when $x=0$ . State the equilibrium solutions and describe the particular solution's long-term behaviour.[9 marks]
Answer
- $y=\frac4{1+3e^{-4x}}$
- Equilibrium solutions $y=0$ and $y=4$ ; the particular solution increases towards $y=4$
Method: The constant solutions lost on division are $y=0$ and $y=4$ . Otherwise separate and integrate to obtain $\frac14(\ln|y|-\ln|4-y|)=x+C$ , so $\frac{y}{4-y}=Ae^{4x}$ . The initial condition gives $A=1/3$ , hence $y=\frac4{1+3e^{-4x}}$ . It starts at $1$ , has positive gradient while $0<y<4$ , and tends to the upper equilibrium $4$ .

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, $v=\frac{\mathrm ds}{\mathrm dt}$ and $a=\frac{\mathrm dv}{\mathrm dt}$ ; 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

1. A population model has solution $N=500e^{0.12t}$ , where $t$ is measured in years. Interpret the constants $500$ and $0.12$ , and state one limitation of the model.[4 marks]
Answer
- Initial population $500$ ; continuous relative growth rate $0.12$ per year
- One valid limitation, such as finite resources eventually preventing unlimited exponential growth
Method: At $t=0$ , $N=500$ , so $500$ is the initial population. Also $\frac1N\frac{\mathrm dN}{\mathrm dt}=0.12$ , so $0.12$ is the continuous proportional growth rate per year. The model predicts unbounded growth and therefore ignores a limiting factor such as finite food, space or changing birth and death rates.

Tier 2 · Standard

1. The velocity of a particle is modelled by $v=20(1-e^{-0.5t})\,\text{m s}^{-1}$ for $t\geq0$ . Find the limiting velocity and the time when $v=15\,\text{m s}^{-1}$ . State one limitation of the model.[6 marks]
Answer
- $20\,\text{m s}^{-1}$
- $t=2\ln4\approx2.77\,\text{s}$
- One valid limitation, such as a constant resistance law being assumed
Method: As $t\to\infty$ , $e^{-0.5t}\to0$ , so the limiting velocity is $20\,\text{m s}^{-1}$ . For $v=15$ , $15=20(1-e^{-0.5t})$ , so $e^{-0.5t}=1/4$ and $t=2\ln4\approx2.77\,\text{s}$ . A real resistance law or driving force may change with conditions, so the same constants need not remain valid indefinitely.

Tier 3 · Hard

1. A particle has velocity $v=18-24e^{-0.3t}\,\text{m s}^{-1}$ for $t\geq0$ , with displacement $s=0$ at $t=0$ . Find when the particle changes direction, its displacement then, and its limiting velocity. Give one limitation of using this model for arbitrarily large $t$ .[8 marks]
Answer
- $t=\frac{10}{3}\ln\frac43\approx0.959\,\text{s}$
- $s\approx-2.74\,\text{m}$
- Limiting velocity $18\,\text{m s}^{-1}$
- One valid limitation tied to the assumed force or resistance law
Method: Solve $v=0$ : $18=24e^{-0.3t}$ , so $t=\frac{10}{3}\ln(4/3)\approx0.959\,\text{s}$ . Since $v(0)=-6$ and $v$ tends to $18>0$ , the continuous velocity changes sign there, proving a reversal. Integrating and using $s(0)=0$ gives $s=18t+80e^{-0.3t}-80$ , which is approximately $-2.74\,\text{m}$ at the turn. The limiting velocity is $18\,\text{m s}^{-1}$ . Over long times, the assumed force or resistance relationship and constant conditions may cease to apply.