Evaluate definite integrals and apply integration to find plane areas between curves and lines

IGCSE Additional Mathematics – Full Syllabus Notes

1. Functions – Refresher

• Definition: A function \(y=f(x)\) assigns exactly one output \(y\) to each permissible input \(x\).
• Domain & Range
  • Find the domain by locating values that make the expression undefined:
    • Division by zero → denominator ≠ 0
    • Even roots → radicand ≥ 0
    • Logarithms → argument > 0
  • Range is the set of possible \(y\)‑values once the domain is fixed.

  Example: \(f(x)=\dfrac{\sqrt{x-2}}{x-5}\)
  • Denominator ≠ 0 ⇒ \(xeq5\)
  • Radicand ≥ 0 ⇒ \(x-2\ge0\) ⇒ \(x\ge2\)
  ⇒ Domain: \([2,5)\cup(5,\infty)\).

• One‑to‑one (injective) functions
  • Horizontal‑line test: every horizontal line cuts the graph at most once.
  • Monotonicity test: if the function is strictly increasing or decreasing on its domain, it is one‑to‑one.

  Example: \(f(x)=x^{2}\) is not one‑to‑one on \(\mathbb R\) (fails the horizontal‑line test). Restricting to \(x\ge0\) makes it strictly increasing, hence one‑to‑one.

• Inverse functions
  1. Ensure the original function is one‑to‑one (restrict the domain if necessary).
  2. Swap \(x\) and \(y\) and solve for the new \(y\).
  3. State the domain of the inverse (which is the range of the original) and its range (the restricted domain).

  Worked example (restricted quadratic): \(y=2x^{2}+3,\;x\ge0\)

  1. Swap: \(x=2y^{2}+3\).
  2. Isolate \(y\): \(y^{2}=\dfrac{x-3}{2}\;\Rightarrow\;y=\sqrt{\dfrac{x-3}{2}}\) (positive root because \(x\ge0\)).
  3. Inverse: \(\displaystyle f^{-1}(x)=\sqrt{\frac{x-3}{2}},\qquad x\ge3.\)

  Graphical note: the graph of an inverse is the reflection of the original graph in the line \(y=x\).

• Composite (combined) functions
  • \((f\circ g)(x)=f\bigl(g(x)\bigr)\) – apply \(g\) first, then \(f\).
  • \((g\circ f)(x)=g\bigl(f(x)\bigr)\) – the order matters.

  Function	Expression
  \(f(x)=2x+1\)
  \(g(x)=x^{2}\)
  \(f\circ g\)	\(f(g(x))=2x^{2}+1\)
  \(g\circ f\)	\(g(f(x))=(2x+1)^{2}=4x^{2}+4x+1\)

2. Quadratic Functions

• Standard form: \(y=ax^{2}+bx+c\) (\(aeq0\)).
• Vertex (completed‑square) form: \(y=a(x-h)^{2}+k\) where \((h,k)\) is the vertex.
• Completing the square (general formula): \[ ax^{2}+bx+c =a\Bigl[\bigl(x+\tfrac{b}{2a}\bigr)^{2} -\tfrac{b^{2}-4ac}{4a^{2}}\Bigr]. \]
• Discriminant: \(\Delta=b^{2}-4ac\)
  • \(\Delta>0\) – two distinct real roots → parabola cuts the \(x\)-axis at two points.
  • \(\Delta=0\) – one repeated real root → parabola is tangent to the \(x\)-axis.
  • \(\Delta<0\) – no real roots → parabola does not meet the \(x\)-axis.
• Maximum / minimum (vertex): \[ x_{\text{v}}=-\frac{b}{2a},\qquad y_{\text{v}}=f\bigl(x_{\text{v}}\bigr)=c-\frac{b^{2}}{4a}. \]

  Using calculus: \(f'(x)=2ax+b\); set \(f'(x)=0\) to obtain the same \(x_{\text{v}}\).

• Sketching a quadratic – 5‑point checklist
  1. Identify \(a\) (opens up if \(a>0\), down if \(a<0\)).
  2. Find the vertex \((h,k)\) using \(-b/2a\) or completing the square.
  3. Determine the axis of symmetry \(x=h\).
  4. Calculate the \(y\)-intercept \((0,c)\).
  5. Locate the real roots (if any) using the discriminant or factorisation.
• Quadratic inequalities
  • Factor the quadratic (or use the sign of \(\Delta\)).
  • Draw a sign‑chart for the factors, marking where the product is \(\ge0\) or \(\le0\).

  Example: Solve \(x^{2}-5x+6\le0\).

  1. Factor: \((x-2)(x-3)\le0\).
  2. Critical points: \(x=2,\;3\).
  3. Sign‑chart:

  Interval	Sign of \((x-2)\)	Sign of \((x-3)\)	Product
  \((-\infty,2)\)	\(-\)	\(-\)	\(+\)
  \([2,3]\)	\(0\) to \(+\)	\(-\) to \(0\)	\(\le0\)
  \((3,\infty)\)	\(+\)	\(+\)	\(+\)

  Solution: \([2,3]\).

3. Factors of Polynomials

• Remainder theorem: When \(P(x)\) is divided by \((x-a)\) the remainder is \(P(a)\).
  Proof sketch: Write \(P(x)=(x-a)Q(x)+R\). Substituting \(x=a\) gives \(P(a)=R\).
• Factor theorem: \((x-a)\) is a factor of \(P(x)\) iff \(P(a)=0\).
• Synthetic division – a quick way to divide by a linear factor and to test possible integer roots.
• Worked example – factorising a cubic: \[ P(x)=x^{3}-6x^{2}+11x-6. \]
  1. Test integer candidates \(\pm1,\pm2,\pm3,\pm6\). \(P(1)=0\) ⇒ \((x-1)\) is a factor.
  2. Synthetic division by \((x-1)\) gives quotient \(x^{2}-5x+6\).
  3. Factor the quadratic: \((x-2)(x-3)\).
  4. Result: \(\displaystyle P(x)=(x-1)(x-2)(x-3).\)

4. Equations & Inequalities (including Moduli)

• Linear equations: solve by isolation or substitution.
• Quadratic equations: factorise, complete the square, or use the quadratic formula \(\displaystyle x=\frac{-b\pm\sqrt{\Delta}}{2a}\).
• Higher‑order equations: factorisation, substitution, or numerical methods (when required by the syllabus).
• Absolute‑value equations \(|ax+b|=c\):
  • If \(c<0\) → no solution.
  • Otherwise solve the two linear equations \(ax+b=c\) and \(ax+b=-c\).
• Absolute‑value inequalities:
  • \(|ax+b|
  • \(|ax+b|>c\) ⇒ \(ax+b-c\) (use “or”).
• Example: Solve \(|2x-3|\le5\).
  1. \(-5\le2x-3\le5\).
  2. Add 3: \(-2\le2x\le8\).
  3. Divide by 2: \(-1\le x\le4\).

5. Simultaneous Equations

• Linear – linear: substitution or elimination.
• Linear – quadratic: substitute the linear expression into the quadratic, solve the resulting quadratic, then back‑substitute.
• Worked example: \[ \begin{cases} y = 2x+1\\[2pt] x^{2}+y^{2}=25 \end{cases} \]
  1. Substitute \(y\): \(x^{2}+(2x+1)^{2}=25\).
  2. Expand: \(x^{2}+4x^{2}+4x+1=25\Rightarrow5x^{2}+4x-24=0\).
  3. Factor: \((5x-6)(x+4)=0\) ⇒ \(x=\dfrac{6}{5}\) or \(x=-4\).
  4. Corresponding \(y\): \(y=2\cdot\dfrac{6}{5}+1=\dfrac{17}{5}\) and \(y=-7\).

6. Logarithmic & Exponential Functions

• Exponential form: \(y=a^{x}\) with \(a>0,\;aeq1\).
• Natural exponential: \(e^{x}\) where \(e\approx2.71828\).
• Logarithm definition: \(y=\log_{a}x \iff a^{y}=x\).
• Key identities
  • \(\log_{a}(bc)=\log_{a}b+\log_{a}c\)
  • \(\log_{a}\!\left(\dfrac{b}{c}\right)=\log_{a}b-\log_{a}c\)
  • \(\log_{a}(b^{k})=k\log_{a}b\)
  • \(\log_{a}x=\dfrac{\ln x}{\ln a}\) (change of base).
• Example – solving an exponential equation: \[ 3^{2x-1}=27. \]
  1. Write RHS as a power of 3: \(27=3^{3}\).
  2. Equate exponents: \(2x-1=3\).
  3. Solve: \(2x=4\Rightarrow x=2.\)

7. Straight‑Line Graphs

• General equation: \(y=mx+c\) where \(m\) is the gradient and \(c\) the \(y\)-intercept.
• Gradient formula: \(m=\dfrac{\Delta y}{\Delta x}\).
• Transformations
  • Vertical shift: \(y=mx+(c+k)\) (up \(k\) units if \(k>0\)).
  • Horizontal shift: \(y=m(x-h)+c\) (right \(h\) units if \(h>0\)).
  • Reflection in the \(x\)-axis: \(y=-mx-c\).
• Example – equation of a line through two points \((2,5)\) and \((6,13)\):
  1. Gradient: \(m=\dfrac{13-5}{6-2}=2\).
  2. Point‑slope form: \(y-5=2(x-2)\) ⇒ \(y=2x+1\).

8. Coordinate Geometry of the Circle

• Standard form: \((x-h)^{2}+(y-k)^{2}=r^{2}\) where \((h,k)\) is the centre and \(r\) the radius.
• General form: \(x^{2}+y^{2}+2gx+2fy+c=0\)
  Centre \((-g,-f)\), radius \(\displaystyle r=\sqrt{g^{2}+f^{2}-c}\) (provided \(g^{2}+f^{2}>c\)).
• Tangent at \((x_{1},y_{1})\) (point lies on the circle): \[ (x_{1}-h)(x-h)+(y_{1}-k)(y-k)=r^{2} \] or, in general form, \[ xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0. \]
• Example – circle through three points \((1,0),\;(0,2),\;(-1,0)\):
  1. By symmetry the centre lies on the \(y\)-axis ⇒ let it be \((0,k)\).
  2. Distance to \((1,0)\): \(1+k^{2}=r^{2}\).
  3. Distance to \((0,2)\): \((2-k)^{2}=r^{2}\).
  4. Equate: \(1+k^{2}=4-4k+k^{2}\) ⇒ \(k=\dfrac34\).
  5. Radius: \(r^{2}=1+\bigl(\tfrac34\bigr)^{2}= \dfrac{25}{16}\) ⇒ \(r=\dfrac54\).
  6. Equation: \((x)^{2}+\bigl(y-\tfrac34\bigr)^{2}= \bigl(\tfrac54\bigr)^{2}.\)

9. Circular Measure

• Radian: the angle subtended by an arc whose length equals the radius.
• Conversion: \[ \theta_{\text{rad}}=\frac{\pi}{180}\,\theta_{\deg},\qquad \theta_{\deg}= \frac{180}{\pi}\,\theta_{\text{rad}}. \]
• Arc length: \(s=r\theta\) (θ in radians).
• Sector area: \(A=\dfrac12 r^{2}\theta\).
• Example – 45° sector of a circle, \(r=7\) cm:
  1. \(\theta=\dfrac{45\pi}{180}= \dfrac{\pi}{4}\) rad.
  2. Arc length: \(s=7\cdot\dfrac{\pi}{4}= \dfrac{7\pi}{4}\) cm.
  3. Sector area: \(A=\tfrac12\cdot7^{2}\cdot\dfrac{\pi}{4}= \dfrac{49\pi}{8}\) cm\(^2\).

10. Trigonometry

Example – solve \(\sin2x=\sqrt3\cos x\) for \(0^{\circ}\le x\le360^{\circ}\)

1. Use \(\sin2x=2\sin x\cos x\): \(2\sin x\cos x=\sqrt3\cos x\).
2. Factor \(\cos x\): \(\cos x\,(2\sin x-\sqrt3)=0\).
3. Case 1: \(\cos x=0\) ⇒ \(x=90^{\circ},\,270^{\circ}\).
4. Case 2: \(2\sin x-\sqrt3=0\) ⇒ \(\sin x=\dfrac{\sqrt3}{2}\) ⇒ \(x=60^{\circ},\,120^{\circ}\).
5. Solution set: \(\{60^{\circ},\,90^{\circ},\,120^{\circ},\,270^{\circ}\}\).

11. Integration – Definite Integrals & Plane Areas

• Antiderivative (indefinite integral): \[ \int f(x)\,dx = F(x)+C\qquad\text{where }F'(x)=f(x). \] Common formulas (to be memorised):
  • \(\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\;(neq-1)\)
  • \(\displaystyle\int \frac{dx}{x}= \ln|x|+C\)
  • \(\displaystyle\int e^{x}\,dx=e^{x}+C\)
  • \(\displaystyle\int a^{x}\,dx=\frac{a^{x}}{\ln a}+C\;(a>0,\;aeq1)\)
  • \(\displaystyle\int \sin x\,dx=-\cos x+C,\quad\int \cos x\,dx=\sin x+C\)
• Definite integral (area under a curve): \[ \int_{a}^{b} f(x)\,dx = \bigl[F(x)\bigr]_{a}^{b}=F(b)-F(a), \] where \(F\) is any antiderivative of \(f\). This is the Fundamental Theorem of Calculus.
• Geometric interpretation
  • If \(f(x)\ge0\) on \([a,b]\) then \(\displaystyle\int_{a}^{b}f(x)\,dx\) equals the signed area between the curve and the \(x\)-axis.
  • If the curve lies below the axis, the integral is negative; the actual (positive) area is \(\bigl|\int_{a}^{b}f(x)\,dx\bigr|\).
• Area between a curve and a straight line
  1. Find the points of intersection (solve \(f(x)=g(x)\)).
  2. Determine which function is on top in each sub‑interval.
  3. Integrate the difference: \[ \text{Area}= \int_{x_{1}}^{x_{2}}\bigl|\,f(x)-g(x)\,\bigr|\,dx =\int_{x_{1}}^{x_{2}}\bigl(f(x)-g(x)\bigr)\,dx \quad\text{(if }f\ge g\text{).} \]
• Area between two curves (no straight line involved): \[ A=\int_{a}^{b}\bigl|\,f(x)-g(x)\,\bigr|\,dx. \] Use the same three‑step procedure as above.
• Typical exam‑style examples

Example 11‑1 – Evaluate a definite integral

Find \(\displaystyle\int_{1}^{4}(3x^{2}-2x+1)\,dx\).

1. Antiderivative: \(F(x)=3\cdot\frac{x^{3}}{3}-2\cdot\frac{x^{2}}{2}+x = x^{3}-x^{2}+x.\)
2. Apply limits: \(F(4)-F(1)=\bigl(4^{3}-4^{2}+4\bigr)-\bigl(1^{3}-1^{2}+1\bigr)=\bigl(64-16+4\bigr)-\bigl(1-1+1\bigr)=52-1=51.\)
3. Result: \(\displaystyle\int_{1}^{4}(3x^{2}-2x+1)\,dx=51.\)

Example 11‑2 – Area between a curve and the \(x\)-axis

Find the area enclosed by \(y=x^{2}-4\) and the \(x\)-axis.

1. Intersection with the axis: set \(x^{2}-4=0\) ⇒ \(x=\pm2.\)
2. Sketch shows the parabola is below the axis on \([-2,2]\); therefore the required area is \[ A=\Bigl|\int_{-2}^{2}(x^{2}-4)\,dx\Bigr|. \]
3. Antiderivative: \(\displaystyle\int(x^{2}-4)\,dx=\frac{x^{3}}{3}-4x.\)
4. Evaluate: \(\bigl[\tfrac{x^{3}}{3}-4x\bigr]_{-2}^{2}= \Bigl(\tfrac{8}{3}-8\Bigr)-\Bigl(-\tfrac{8}{3}+8\Bigr)=\Bigl(-\tfrac{16}{3}\Bigr)-\Bigl(\tfrac{16}{3}\Bigr)=-\tfrac{32}{3}.\)
5. Take absolute value: \(A=\dfrac{32}{3}\) square units.

Example 11‑3 – Area between a curve and a straight line

Find the area enclosed by the parabola \(y= x^{2}\) and the line \(y=2x+3\).

1. Intersection points: solve \(x^{2}=2x+3\) ⇒ \(x^{2}-2x-3=0\) ⇒ \((x-3)(x+1)=0\) ⇒ \(x=-1,\;3.\)
2. For \(-1\le x\le3\) the line lies above the parabola (check at \(x=0\): \(0<3\)).
3. Area: \[ A=\int_{-1}^{3}\bigl[(2x+3)-x^{2}\bigr]\,dx. \]
4. Antiderivative: \(\displaystyle\int(2x+3-x^{2})dx = x^{2}+3x-\frac{x^{3}}{3}.\)
5. Evaluate: \[ \Bigl[x^{2}+3x-\frac{x^{3}}{3}\Bigr]_{-1}^{3} =\Bigl(9+9-9\Bigr)-\Bigl(1-3+\frac{1}{3}\Bigr) =9-(-\tfrac{5}{3})= \frac{32}{3}. \]
6. Result: \(A=\dfrac{32}{3}\) square units.

Example 11‑4 – Area between two curves (different functions of \(x\) and \(y\))

Find the area bounded by \(y=\sqrt{x}\), the line \(y=0\) and the vertical line \(x=4\).

1. The region is under the curve \(\sqrt{x}\) from \(x=0\) to \(x=4\).
2. Integral: \(A=\displaystyle\int_{0}^{4}\sqrt{x}\,dx=\int_{0}^{4}x^{1/2}dx.\)
3. Antiderivative: \(\displaystyle\frac{x^{3/2}}{3/2}= \frac{2}{3}x^{3/2}.\)
4. Evaluate: \(\frac{2}{3}\bigl[4^{3/2}-0\bigr]=\frac{2}{3}\times8= \frac{16}{3}\) square units.

Key steps to remember for any area problem

1. Identify the bounding curves/lines.
2. Find all points of intersection (solve equations).
3. Decide which curve is upper (or left/right for \(y\)-integrals).
4. Set up the definite integral of the difference (top – bottom).
5. Integrate, apply limits, and take absolute values if the integrand becomes negative.

These notes cover the core IGCSE Additional Mathematics syllabus, present the required methods in a clear, exam‑focused order, and include worked examples that illustrate each technique. Use them for revision, practice, and as a quick reference during examinations.