Integration: techniques, definite integrals, areas under curves

Integration: Techniques, Definite Integrals, Areas & Volumes

1. Why Study Integration?

• It is the inverse operation of differentiation – the antiderivative of a function.
• It gives accumulated quantities (distance, work, charge, etc.).
• It provides signed areas under a curve and geometric areas/volumes between curves.

2. Basic Antiderivatives (Syllabus 1.8.1)

For a function $f(x)$, an antiderivative $F(x)$ satisfies $F'(x)=f(x)$. The most frequently used forms are:

Integral	Result
\displaystyle\int k\,dx	$kx+C$ ($k$ constant)
\displaystyle\int x^{n}\,dx	$\displaystyle\frac{x^{\,n+1}}{n+1}+C\quad(neq-1)$
\displaystyle\int e^{ax}\,dx	$\displaystyle\frac{1}{a}e^{ax}+C$
\displaystyle\int \sin(ax)\,dx	$-\displaystyle\frac{1}{a}\cos(ax)+C$
\displaystyle\int \cos(ax)\,dx	$\displaystyle\frac{1}{a}\sin(ax)+C$

Constant of Integration – Why It Matters

The symbol $C$ represents an arbitrary constant. It is essential when a differential equation is solved, because the constant is fixed by an initial condition.

3. Integration Techniques (Syllabus 1.8.2)

Technique	When to Use	Key Steps
Substitution (u‑substitution)	Integrand contains a function and its derivative, i.e. $f(g(x))g'(x)$.	Set $u=g(x)$. Compute $du=g'(x)\,dx$ and rewrite $dx$. Replace the whole integral by $\int f(u)\,du$. Integrate, then substitute $u=g(x)$ back.
Integration by Parts	Product of two functions where one simplifies on differentiation (e.g. $u\,v'$).	Choose $u$ (so that $du$ is simpler) and $dv$. Find $du$ and $v=\int dv$. Apply $\displaystyle\int u\,dv = uv-\int v\,du$.
Partial Fractions – Linear Factors (Paper 1)	Rational function whose denominator splits into distinct linear factors.	Express $\displaystyle\frac{P(x)}{(x-a_1)(x-a_2)\dots}= \frac{A_1}{x-a_1}+ \frac{A_2}{x-a_2}+ \dots$. Determine the constants $A_i$ (cover‑up method or equating coefficients). Integrate each simple term $\displaystyle\int\frac{A_i}{x-a_i}\,dx =A_i\ln|x-a_i|+C$. Note: Decomposition involving irreducible quadratics belongs to Paper 3 (Pure Mathematics 2) and is not required for Paper 1.
Trigonometric Integrals	Integrands containing powers of $\sin x$, $\cos x$, or products thereof.	Use identities: $\sin^{2}x=\frac{1-\cos2x}{2}$, $\cos^{2}x=\frac{1+\cos2x}{2}$, $\sin x\cos x=\frac{1}{2}\sin2x$. Reduce the powers until a standard integral is reached.
Trigonometric Substitution (Paper 2)	Integrals containing $\sqrt{a^{2}-x^{2}}$, $\sqrt{a^{2}+x^{2}}$, or $\sqrt{x^{2}-a^{2}}$.	Choose the appropriate substitution: $x=a\sin\theta$ for $\sqrt{a^{2}-x^{2}}$ $x=a\tan\theta$ for $\sqrt{a^{2}+x^{2}}$ $x=a\sec\theta$ for $\sqrt{x^{2}-a^{2}}$ Rewrite $dx$ and the radical in terms of $\theta$. Integrate the resulting trigonometric integral. Return to $x$ using a right‑triangle diagram. Example (Paper 2 style) $\displaystyle\int\frac{dx}{\sqrt{a^{2}-x^{2}}}$ Set $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$, $\sqrt{a^{2}-x^{2}}=a\cos\theta$. The integral becomes $\displaystyle\int\frac{a\cos\theta\,d\theta}{a\cos\theta}= \int d\theta =\theta+C$. Since $x=a\sin\theta$, $\theta=\arcsin\!\frac{x}{a}$, so the antiderivative is $\displaystyle\arcsin\!\frac{x}{a}+C$.
Integration of Rational Functions with Irreducible Quadratics (Paper 3 – note)	When the denominator contains a quadratic that cannot be factored over the reals.	These require completing the square and a combination of the previous techniques (partial fractions with linear numerator, plus a trig‑substitution or arctangent form). It is beyond Paper 1, but students should be aware that the syllabus later expects: $\displaystyle\int\frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\arctan\frac{x}{a}+C$ $\displaystyle\int\frac{x\,dx}{x^{2}+a^{2}} = \frac{1}{2}\ln|x^{2}+a^{2}|+C$ These formulas are listed for reference only.

4. Improper Integrals (Syllabus 1.8.3)

An integral is **improper** if

• the integrand becomes unbounded at an endpoint, or
• the interval of integration is infinite.

It is defined as a limit. The Cambridge syllabus expects the following elementary cases.

4.1 Convergent Example (integrand singular at a finite endpoint)

4.2 Divergent Example (logarithmic singularity)

4.3 Improper Integral on an Infinite Interval

5. Definite Integrals (Syllabus 1.8.4)

5.1 Fundamental Theorem of Calculus (FTC)

If $F$ is an antiderivative of $f$ on $[a,b]$, then

\[ \int_{a}^{b}f(x)\,dx = F(b)-F(a). \]

5.2 Useful Properties

$\displaystyle\int_{a}^{a} f(x)\,dx = 0$

$\displaystyle\int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx$

$\displaystyle\int_{a}^{b} [f(x)\pm g(x)]\,dx = \int_{a}^{b} f(x)\,dx \pm \int_{a}^{b} g(x)\,dx$

$\displaystyle\int_{a}^{b} k\,f(x)\,dx = k\int_{a}^{b} f(x)\,dx\qquad(k\text{ constant})$

6. Areas and Volumes (Syllabus 1.8.5)

6.1 Area under a curve (with respect to the $x$‑axis)

If $f(x)\ge0$ on $[a,b]$, the geometric area is

\[ \text{Area}= \int_{a}^{b} f(x)\,dx . \]

If $f$ changes sign, split the interval at the zeros and add the absolute values of the resulting signed areas.

6.2 Area between two curves $y=f(x)$ and $y=g(x)$

Assume $f(x)\ge g(x)$ on $[a,b]$.

\[ \text{Area}= \int_{a}^{b}\bigl[f(x)-g(x)\bigr]\,dx . \]

6.3 Area expressed with $y$ as the variable

When the region is more naturally described by $x=h(y)$ and $x=k(y)$ (with $h(y)\ge k(y)$ on $[c,d]$), use

\[ \text{Area}= \int_{c}^{d}\bigl[h(y)-k(y)\bigr]\,dy . \]

6.4 Volumes of Revolution (Paper 1 – basic cases)

• Disk/Washer method (rotation about the $x$‑axis) \[ V = \pi\int_{a}^{b}\bigl[\,R(x)^{2}-r(x)^{2}\,\bigr]\,dx, \] where $R$ is the outer radius and $r$ the inner radius (if a hole is present).
• Shell method (rotation about the $y$‑axis) \[ V = 2\pi\int_{c}^{d} \bigl[\,\text{radius}\times\text{height}\,\bigr]\,dy. \] For rotation about the $y$‑axis, the radius is $x$ (or $|x|$) and the height is $f(x)-g(x)$ expressed as a function of $x$.

Example – Volume by the Disk Method

Find the volume generated by rotating the region bounded by $y=x^{2}$ and $y=2x$ about the $x$‑axis.

1. Intersection points: $x^{2}=2x\;\Rightarrow\;x=0,\,2$.
2. For $0\le x\le2$, the upper curve is $y=2x$, the lower is $y=x^{2}$.
3. Outer radius $R(x)=2x$, inner radius $r(x)=x^{2}$.
4. Volume: \[ V=\pi\int_{0}^{2}\!\bigl[(2x)^{2}-(x^{2})^{2}\bigr]dx =\pi\int_{0}^{2}\!(4x^{2}-x^{4})dx. \]
5. Integrate: $\displaystyle\Bigl[\frac{4}{3}x^{3}-\frac{x^{5}}{5}\Bigr]_{0}^{2} =\frac{32}{3}-\frac{32}{5}= \frac{96-64}{15}= \frac{32}{15}$.
6. Hence $V=\displaystyle\frac{32\pi}{15}\,$ cubic units.

7. Worked Examples (All Paper 1 Techniques)

Example 1 – Substitution

Evaluate $\displaystyle\int (3x^{2}+2)\,e^{x^{3}+2x}\,dx$.

1. Let $u=x^{3}+2x\;\Rightarrow\;du=(3x^{2}+2)\,dx$.
2. Integral becomes $\displaystyle\int e^{u}\,du = e^{u}+C$.
3. Result: $e^{x^{3}+2x}+C$.

Example 2 – Integration by Parts

Find $\displaystyle\int x\cos x\,dx$.

1. $u=x\;\Rightarrow\;du=dx$.
2. $dv=\cos x\,dx\;\Rightarrow\;v=\sin x$.
3. Apply the formula: $\displaystyle\int x\cos x\,dx = x\sin x-\int\sin x\,dx$.
4. Integrate the remaining term: $\int\sin x\,dx = -\cos x$.
5. Result: $x\sin x+\cos x+C$.

Example 3 – Partial Fractions (Linear Factors)

Compute $\displaystyle\int\frac{2x+5}{(x+1)(x+2)}\,dx$.

1. Decompose: $\displaystyle\frac{2x+5}{(x+1)(x+2)} = \frac{A}{x+1}+\frac{B}{x+2}$.
2. Multiply through: $2x+5=A(x+2)+B(x+1)$.
3. Set $x=-1$: $2(-1)+5=A(1)\Rightarrow A=3$.
4. Set $x=-2$: $2(-2)+5=B(-1)\Rightarrow B=-1$.
5. Integral: $\displaystyle\int\!\Bigl(\frac{3}{x+1}-\frac{1}{x+2}\Bigr)dx =3\ln|x+1|-\ln|x+2|+C$.

Example 4 – Trigonometric Integral

Evaluate $\displaystyle\int \sin^{3}x\cos x\,dx$.

1. Write $\sin^{3}x=(1-\cos^{2}x)\sin x$.
2. Set $u=\cos x\;\Rightarrow\;du=-\sin x\,dx$.
3. Integral becomes $-\displaystyle\int(1-u^{2})\,du =-\bigl(u-\tfrac{u^{3}}{3}\bigr)+C$.
4. Back‑substitute: $-\cos x+\frac{\cos^{3}x}{3}+C$.

Example 5 – Area Between Two Curves

Find the area bounded by $y=x^{2}$ and $y=2x$.

1. Intersection: $x^{2}=2x\Rightarrow x=0,\,2$.
2. On $[0,2]$, $2x\ge x^{2}$.
3. Area $= \displaystyle\int_{0}^{2}(2x-x^{2})\,dx$.
4. Antiderivative: $x^{2}-\dfrac{x^{3}}{3}$.
5. Evaluate: $\bigl[4-\tfrac{8}{3}\bigr]-0 = \dfrac{4}{3}$ square units.

Example 6 – Improper Integral (Convergent on an Infinite Interval)

Evaluate $\displaystyle\int_{1}^{\infty}\frac{dx}{x^{2}}$.

1. Write as $\displaystyle\lim_{b\to\infty}\int_{1}^{b}x^{-2}\,dx$.
2. Antiderivative: $-x^{-1}$.
3. Limit: $\displaystyle\lim_{b\to\infty}\bigl[-b^{-1}+1\bigr]=1$ (convergent).

8. Summary Checklist (Paper 1)

• Identify the most suitable technique before starting an integration.
• For substitution, always rewrite $dx$ in terms of $du$.
• For integration by parts, pick $u$ so that $du$ is simpler than $u$.
• Partial‑fraction decomposition is limited to distinct linear factors in Paper 1.
• Remember the constant of integration $C$ – it is required when a particular solution is sought.
• Recognise improper integrals (unbounded integrand or infinite limits) and evaluate them as limits.
• Apply the Fundamental Theorem of Calculus after finding an antiderivative for any definite integral.
• When computing areas:
  • Split the interval at any zeros of the function.
  • Subtract the lower curve from the upper curve on each sub‑interval.
  • If the region is described by $x$ as a function of $y$, integrate with respect to $y$.
• For volumes of revolution, decide whether the disk/washer or shell method gives the simpler integral.