Integration

Explanation

• 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 $c$.

Explanation

• For $n\ne-1$, $\int xⁿ\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; 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 $kx$.
• Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$.

Explanation

• 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.
• Intersections satisfy $4x-x^2=x$, so $x(3-x)=0$ and $x=0,3$.
• For a parametric curve, its signed area is $\int y\,\dfrac{dx}{dt}\,dt$ over the parameter interval, with limits ordered to match the traced direction.

Explanation

• 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.

Explanation

• 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.
• A required standard result is $\int\ln x\,dx=x\ln x-x+c$, obtained by integration by parts with $u=\ln x$ and $dv=dx$.

Explanation

• 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.
• Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$.
• Then $5x+1=A(x+2)+B(x-1)$.

Explanation

• 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.

Explanation

• 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.