5.5, 5.10

Indefinite Integration & The Reverse Power Rule

  • Antidifferentiation Concept: Integration is the inverse process of differentiation. If $\frac{d}{dx}[F(x)] = f(x)$, then: $$\int f(x) \, dx = F(x) + C$$ where $C \in \mathbb{R}$ is the arbitrary constant of integration. Omitting $+C$ on an indefinite integral loses accuracy marks on IB Paper 1.
  • Reverse Power Rule: Add 1 to the exponent and divide by the new exponent: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$
  • Algebraic Preparation Before Integrating:
    • Expand products: $\int (2x - 3)(x + 4) \, dx = \int (2x^2 + 5x - 12) \, dx$. Never integrate factors individually!
    • Split fractions with a single monomial denominator: $$\int \frac{3x^3 - 5x + 2}{x^2} \, dx = \int \left( 3x - 5x^{-1} + 2x^{-2} \right) dx$$
    • Rewrite radicals into fractional exponents: $\sqrt{x} = x^{1/2}$, $\frac{1}{\sqrt[3]{x^2}} = x^{-2/3}$.
  • Finding the Constant $C$ (Boundary Condition): Given $f'(x)$ and a known coordinate point $(x_0, y_0)$ on the curve, substitute $x_0$ and $y_0 = f(x_0)$ to solve for the unique value of $C$.

Example 1

Finding the Original Curve from a Derivative

The gradient of a curve is given by $f'(x) = 6x^2 - \frac{4}{\sqrt{x}} + 3$ for $x > 0$. Given that the curve passes through the point $(4, 115)$, find $f(x)$.

  1. Rewrite into index form: $$f'(x) = 6x^2 - 4x^{-1/2} + 3$$
  2. Integrate term by term: $$f(x) = \int \left( 6x^2 - 4x^{-1/2} + 3 \right) dx = 6\left(\frac{x^3}{3}\right) - 4\left(\frac{x^{1/2}}{1/2}\right) + 3x + C$$ $$f(x) = 2x^3 - 8\sqrt{x} + 3x + C$$
  3. Substitute the boundary point $(4, 115)$ to solve for $C$: $$115 = 2(4)^3 - 8\sqrt{4} + 3(4) + C$$ $$115 = 2(64) - 8(2) + 12 + C$$ $$115 = 128 - 16 + 12 + C = 124 + C \implies C = 115 - 124 = -9$$
  4. State the final equation of the curve: $$f(x) = 2x^3 - 8\sqrt{x} + 3x - 9$$

Answer: $f(x) = 2x^3 - 8\sqrt{x} + 3x - 9$

5.10

Standard Transcendental Integrals & Linear Composites

  • Standard Functions (Formula Booklet): $$\int e^x \, dx = e^x + C, \qquad \int \frac{1}{x} \, dx = \ln|x| + C \quad (x \neq 0)$$ $$\int \cos x \, dx = \sin x + C, \qquad \int \sin x \, dx = -\cos x + C \quad (\text{Note the negative!})$$
  • Linear Composite Rule (Inspection / Reverse Chain Rule): When the inner argument is a linear function $ax + b$, divide the antiderivative by the coefficient $a$: $$\int f(ax + b) \, dx = \frac{1}{a} F(ax + b) + C$$ Specific cases frequently assessed on IB exams: $$\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C$$ $$\int \frac{1}{ax + b} \, dx = \frac{1}{a}\ln|ax + b| + C$$ $$\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C, \qquad \int \sin(kx) \, dx = -\frac{1}{k}\cos(kx) + C$$ $$\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C \quad (n \neq -1)$$
  • Limitation: This division by $a$ ONLY works for linear inner expressions. For arguments like $(x^2 + 1)^3$ or $e^{x^2}$, you must use formal substitution!

Example 2

Integrating Composite Expressions by Inspection

Evaluate the indefinite integral $\int \left( 4e^{-3x} + \frac{6}{2x + 5} - 5\sin(2x) \right) dx$.

  1. Integrate the exponential term $\int 4e^{-3x} \, dx$: $$\text{Linear argument } -3x \implies \text{divide by } -3: \quad 4\left(\frac{e^{-3x}}{-3}\right) = -\frac{4}{3}e^{-3x}$$
  2. Integrate the reciprocal term $\int \frac{6}{2x + 5} \, dx$: $$\text{Linear argument } 2x + 5 \implies \text{divide by } 2: \quad 6 \cdot \frac{1}{2} \ln|2x + 5| = 3\ln|2x + 5|$$
  3. Integrate the trigonometric term $\int -5\sin(2x) \, dx$: $$\int \sin(2x) \, dx = -\frac{1}{2}\cos(2x) \implies -5\left(-\frac{1}{2}\cos(2x)\right) = +\frac{5}{2}\cos(2x)$$
  4. Combine all terms and append $+C$: $$-\frac{4}{3}e^{-3x} + 3\ln|2x + 5| + \frac{5}{2}\cos(2x) + C$$

Answer: $-\frac{4}{3}e^{-3x} + 3\ln|2x + 5| + \frac{5}{2}\cos(2x) + C$

5.10

Integration by Substitution (Reverse Chain Rule)

  • Pattern Recognition: Used when the integrand contains a composite function multiplied by a scalar multiple of its inner derivative: $$\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)$$
  • Fractional Pattern (Logarithmic Derivative): $$\int \frac{g'(x)}{g(x)} \, dx = \ln|g(x)| + C$$
  • The 4-Step Systematic Substitution Method:
    1. Select $u$: Choose the inside function (under a power, in an exponent, inside trig, or in the denominator).
    2. Differentiate to express $dx$: Compute $\frac{du}{dx} = g'(x) \implies dx = \frac{du}{g'(x)}$.
    3. Substitute into the integral: Replace $g(x)$ with $u$ and $dx$ with $\frac{du}{g'(x)}$. The remaining $x$-terms must cancel cleanly.
    4. Evaluate and convert:
      • For indefinite integrals: evaluate in $u$, then substitute back $u = g(x)$.
      • For definite integrals: convert the $x$-limits to $u$-limits ($u_a = g(a)$, $u_b = g(b)$) and evaluate directly in $u$. Never mix $x$ and $u$ variables!

Example 3

Definite Integration with Variable Transformation

Evaluate the definite integral $\int_{0}^{1} 6x(x^2 + 3)^2 \, dx$ by changing the integration limits.

  1. Choose $u$ and compute the differential: $$u = x^2 + 3 \implies \frac{du}{dx} = 2x \implies dx = \frac{du}{2x}$$
  2. Transform the limits of integration: $$\text{Lower limit: } x = 0 \implies u = 0^2 + 3 = 3$$ $$\text{Upper limit: } x = 1 \implies u = 1^2 + 3 = 4$$
  3. Substitute into the integral and cancel $x$: $$\int_{0}^{1} 6x(x^2 + 3)^2 \, dx = \int_{3}^{4} 6x \cdot u^2 \cdot \frac{du}{2x} = \int_{3}^{4} 3u^2 \, du$$
  4. Evaluate the transformed definite integral directly: $$\int_{3}^{4} 3u^2 \, du = [u^3]_{3}^{4} = 4^3 - 3^3 = 64 - 27 = 37$$

Answer: $37$

5.5, 5.11

Definite Integrals, Signed Area & Crossing the $x$-Axis

  • The Fundamental Theorem of Calculus: $$\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)$$ where $F'(x) = f(x)$. No constant $+C$ is needed for definite integrals because $C - C = 0$.
  • Signed Integral vs. Geometric Area:
    • A definite integral measures net signed area: regions where $f(x) > 0$ contribute positively, while regions where $f(x) < 0$ contribute negatively.
    • Geometric area is strictly non-negative: $$\text{Total Geometric Area} = \int_a^b |f(x)| \, dx$$
  • Protocol When a Curve Crosses the $x$-Axis on $[a, b]$:
    1. Find all $x$-intercepts by solving $f(x) = 0$.
    2. Identify which roots lie within the interval $[a, b]$.
    3. Split the integral at each root: $$\text{Area} = \int_a^c f(x) \, dx + \left| \int_c^b f(x) \, dx \right| = \int_a^c f(x) \, dx - \int_c^b f(x) \, dx$$
Signed Definite Integral vs Total Geometric Area Shows a curve crossing the x-axis at x = c between limits a and b, highlighting positive area A1 above the axis and negative area A2 below the axis. x y y = f(x) x = a x = c x = b A₁ > 0 A₂ < 0 Net Integral: ∫ₐᵇ f(x)dx = A₁ − |A₂| Total Area: A₁ + |A₂| = ∫ₐᶜ f(x)dx − ∫꜀ᵇ f(x)dx
Figure 5.7: Total geometric area requires splitting the integral at the $x$-intercept $c$, taking the absolute value of the negative region below the axis.

Example 4

Calculating Total Geometric Area with an Axis Crossing

Find the total geometric area enclosed by the curve $y = x^2 - 4x + 3$ and the $x$-axis between $x = 0$ and $x = 3$.

  1. Find $x$-intercepts by factoring $f(x) = 0$: $$x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \implies x = 1 \quad \text{and} \quad x = 3$$ The root $x = 1$ splits the domain $[0, 3]$ into two distinct intervals: $[0, 1]$ and $[1, 3]$.
  2. Evaluate the antiderivative function $F(x)$: $$F(x) = \int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x$$
  3. Evaluate the first region $[0, 1]$ (above $x$-axis): $$A_1 = \int_0^1 (x^2 - 4x + 3) \, dx = F(1) - F(0) = \left(\frac{1}{3} - 2 + 3\right) - 0 = \frac{4}{3}$$
  4. Evaluate the second region $[1, 3]$ (below $x$-axis): $$\int_1^3 (x^2 - 4x + 3) \, dx = F(3) - F(1)$$ $$F(3) = \frac{3^3}{3} - 2(3^2) + 3(3) = 9 - 18 + 9 = 0$$ $$\int_1^3 (x^2 - 4x + 3) \, dx = 0 - \frac{4}{3} = -\frac{4}{3}$$ Because area is physical, take the absolute value: $A_2 = \left|-\frac{4}{3}\right| = \frac{4}{3}$.
  5. Add the two regions together: $$\text{Total Area} = A_1 + A_2 = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$$

Answer: $\frac{8}{3}$ square units

5.11

Area Enclosed Between Two Intersecting Curves

  • Fundamental Principle: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$, is: $$\text{Area} = \int_a^b \left[ y_{\text{upper}} - y_{\text{lower}} \right] \, dx = \int_a^b \left[ f(x) - g(x) \right] \, dx$$
  • Why the Relative Axis Position Does Not Matter: Because $[f(x) - g(x)]$ represents the positive vertical distance between curves, this formula holds true whether the region is above, below, or straddling the $x$-axis!
  • 3-Step Method for Enclosed Areas:
    1. Find intersection points: Set $f(x) = g(x)$ and solve for the roots $x = a$ and $x = b$. These roots form the limits of integration.
    2. Determine upper vs. lower curve: Test a test value $x_t \in (a, b)$. If $f(x_t) > g(x_t)$, then $f(x)$ is the upper function.
    3. Integrate the difference: Evaluate $\int_a^b [f(x) - g(x)] \, dx$.
Area Enclosed Between Two Curves Shows an upper concave-down parabola and a lower straight line intersecting at x = a and x = b, with the region between them shaded. x y y = g(x) y = f(x) (Upper) h = f − g dx A(a, g(a)) x = a B(b, g(b)) x = b Area = ∫ₐᵇ [f(x) − g(x)] dx
Figure 5.8: The area bounded between two curves equals the integral of the upper curve minus the lower curve between their intersection coordinates $a$ and $b$.

Example 5

Finding the Area Between a Parabola and a Line

Find the area of the finite region enclosed between the curve $y = 8 - x^2$ and the line $y = x + 2$.

  1. Find the intersection points by equating equations: $$8 - x^2 = x + 2 \implies x^2 + x - 6 = 0$$ $$(x + 3)(x - 2) = 0 \implies x = -3 \quad \text{and} \quad x = 2$$ The limits of integration are $a = -3$ and $b = 2$.
  2. Identify which function is on top: $$\text{Test } x = 0: \quad y_{\text{curve}} = 8 - 0 = 8, \quad y_{\text{line}} = 0 + 2 = 2$$ Since $8 > 2$, the parabola $y = 8 - x^2$ is the upper function.
  3. Set up the integral: $$\text{Area} = \int_{-3}^{2} \left[ (8 - x^2) - (x + 2) \right] dx = \int_{-3}^{2} \left( 6 - x - x^2 \right) dx$$
  4. Evaluate the definite integral: $$\int_{-3}^{2} (6 - x - x^2) \, dx = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2}$$ $$\text{At } x = 2: \quad 6(2) - \frac{2^2}{2} - \frac{2^3}{3} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$ $$\text{At } x = -3: \quad 6(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{3} = -18 - \frac{9}{2} + 9 = -9 - \frac{9}{2} = -\frac{27}{2}$$ $$\text{Area} = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{44 + 81}{6} = \frac{125}{6}$$

Answer: $\frac{125}{6}$ square units ($\approx 20.83$)

Tier 1: Core Fluency (Questions 1–4)
Exercise 1 Paper 1 Core Skill

Find the indefinite integral:

$$\int \left( 6x^2 - \frac{5}{\sqrt{x}} + \frac{2}{x^3} \right) dx$$

View Complete Solution
  1. Convert all fractional and radical terms to index form: $$\int \left( 6x^2 - 5x^{-1/2} + 2x^{-3} \right) dx$$
  2. Apply the reverse power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$: $$= 6\left(\frac{x^3}{3}\right) - 5\left(\frac{x^{1/2}}{1/2}\right) + 2\left(\frac{x^{-2}}{-2}\right) + C$$
  3. Simplify the coefficients: $$= 2x^3 - 10x^{1/2} - x^{-2} + C = 2x^3 - 10\sqrt{x} - \frac{1}{x^2} + C$$

Final Answer: $2x^3 - 10\sqrt{x} - \frac{1}{x^2} + C$

Exercise 2 Paper 1 Initial Value Problem

A curve has gradient function $\frac{dy}{dx} = 4e^{2x} + 3\cos x$ and passes through the point $(0, 7)$. Find the exact equation of the curve in the form $y = f(x)$.

View Complete Solution
  1. Integrate the gradient function using linear composite rules: $$y = \int \left( 4e^{2x} + 3\cos x \right) dx = 4\left(\frac{e^{2x}}{2}\right) + 3\sin x + C = 2e^{2x} + 3\sin x + C$$
  2. Substitute the initial point $(0, 7)$ to find $C$: $$7 = 2e^{2(0)} + 3\sin(0) + C$$ $$7 = 2(1) + 0 + C \implies C = 5$$
  3. Write the final equation: $$y = 2e^{2x} + 3\sin x + 5$$

Final Answer: $y = 2e^{2x} + 3\sin x + 5$

Exercise 3 Paper 1 Linear Composite Evaluation

Evaluate the definite integral:

$$\int_{0}^{3} (2x - 1)^3 \, dx$$

View Complete Solution
  1. Use the linear power rule $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}$ with $a = 2, n = 3$: $$\int (2x - 1)^3 \, dx = \left[ \frac{(2x - 1)^4}{2 \cdot 4} \right]_{0}^{3} = \left[ \frac{(2x - 1)^4}{8} \right]_{0}^{3}$$
  2. Evaluate at the upper limit $x = 3$: $$\frac{(2(3) - 1)^4}{8} = \frac{5^4}{8} = \frac{625}{8}$$
  3. Evaluate at the lower limit $x = 0$: $$\frac{(2(0) - 1)^4}{8} = \frac{(-1)^4}{8} = \frac{1}{8}$$
  4. Subtract the limits: $$\frac{625}{8} - \frac{1}{8} = \frac{624}{8} = 78$$

Final Answer: $78$

Exercise 4 Paper 1 Substitution with Limits

Evaluate the definite integral by substitution:

$$\int_{0}^{\sqrt{\pi}} 2x \sin(x^2) \, dx$$

View Complete Solution
  1. Choose $u = x^2$: $$\frac{du}{dx} = 2x \implies dx = \frac{du}{2x}$$
  2. Transform the limits: $$x = 0 \implies u = 0^2 = 0$$ $$x = \sqrt{\pi} \implies u = (\sqrt{\pi})^2 = \pi$$
  3. Substitute into the integral: $$\int_{0}^{\pi} 2x \sin(u) \cdot \frac{du}{2x} = \int_{0}^{\pi} \sin u \, du$$
  4. Integrate and evaluate: $$[-\cos u]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$$

Final Answer: $2$

Tier 2: Exam Modeling (Questions 5–7)
Exercise 5 Paper 1 Geometric Area Split

Find the total geometric area bounded by the curve $y = x^2 - 3x$ and the $x$-axis between $x = 0$ and $x = 4$.

View Complete Solution
  1. Find roots of the curve: $$x^2 - 3x = x(x - 3) = 0 \implies x = 0 \quad \text{and} \quad x = 3$$ The curve lies below the $x$-axis on $[0, 3]$ and above the $x$-axis on $[3, 4]$.
  2. Evaluate the antiderivative $F(x) = \frac{x^3}{3} - \frac{3x^2}{2}$:
  3. Evaluate Region 1 ($[0, 3]$): $$\int_{0}^{3} (x^2 - 3x) \, dx = \left[ \frac{3^3}{3} - \frac{3(9)}{2} \right] - 0 = 9 - \frac{27}{2} = -\frac{9}{2}$$ $$\text{Area}_1 = \left|-\frac{9}{2}\right| = \frac{9}{2}$$
  4. Evaluate Region 2 ($[3, 4]$): $$\int_{3}^{4} (x^2 - 3x) \, dx = F(4) - F(3) = \left(\frac{64}{3} - 24\right) - \left(-\frac{9}{2}\right) = -\frac{8}{3} + \frac{9}{2} = \frac{11}{6}$$ $$\text{Area}_2 = \frac{11}{6}$$
  5. Sum the geometric areas: $$\text{Total Area} = \frac{9}{2} + \frac{11}{6} = \frac{27 + 11}{6} = \frac{38}{6} = \frac{19}{3}$$

Final Answer: $\frac{19}{3}$ square units ($\approx 6.33$)

Exercise 6 Paper 1 / Paper 2 Area Between Graphs

Find the area of the region completely enclosed between the parabola $y = 5 - x^2$ and the line $y = x - 1$.

View Complete Solution
  1. Find intersection points: $$5 - x^2 = x - 1 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0$$ $$x = -3 \quad \text{and} \quad x = 2$$
  2. Check which curve is upper: $$\text{At } x = 0: \quad y = 5 - 0^2 = 5, \quad y = 0 - 1 = -1 \implies \text{Parabola is upper.}$$
  3. Integrate $(y_{\text{upper}} - y_{\text{lower}})$: $$\text{Area} = \int_{-3}^{2} \left[ (5 - x^2) - (x - 1) \right] dx = \int_{-3}^{2} (6 - x - x^2) \, dx$$ $$= \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2}$$
  4. Evaluate limits: $$\text{Upper limit } (x=2): \quad 12 - 2 - \frac{8}{3} = \frac{22}{3}$$ $$\text{Lower limit } (x=-3): \quad -18 - \frac{9}{2} + 9 = -\frac{27}{2}$$ $$\text{Area} = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{125}{6}$$

Final Answer: $\frac{125}{6}$ square units

Exercise 7 Paper 1 Logarithmic Derivative Form

Evaluate the definite integral:

$$\int_{0}^{2} \frac{4x}{x^2 + 1} \, dx$$

Express your answer in the form $\ln k$, where $k \in \mathbb{Z}^+$.

View Complete Solution
  1. Recognize the $\frac{g'(x)}{g(x)}$ pattern: $$\text{Let } u = x^2 + 1 \implies \frac{du}{dx} = 2x \implies 4x \, dx = 2 \, du$$
  2. Transform limits: $$x = 0 \implies u = 0^2 + 1 = 1$$ $$x = 2 \implies u = 2^2 + 1 = 5$$
  3. Substitute and integrate: $$\int_{1}^{5} \frac{2}{u} \, du = 2[\ln|u|]_{1}^{5} = 2(\ln 5 - \ln 1) = 2\ln 5$$
  4. Use logarithm power property $a\ln b = \ln(b^a)$: $$2\ln 5 = \ln(5^2) = \ln 25$$

Final Answer: $\ln 25$ (hence $k = 25$)

Tier 3: Complex Applications & IB Exam Modeling (Questions 8–10)
Exercise 8 Paper 1 Parameter Limit Problem

The area under the curve $y = \frac{12}{x^2}$ from $x = 2$ to $x = k$ (where $k > 2$) is equal to $4$. Find the exact value of $k$.

View Complete Solution
  1. Set up the area integral equation: $$\int_{2}^{k} 12x^{-2} \, dx = 4$$
  2. Integrate $12x^{-2}$: $$\left[ -\frac{12}{x} \right]_{2}^{k} = 4$$
  3. Substitute limits: $$\left( -\frac{12}{k} \right) - \left( -\frac{12}{2} \right) = 4$$ $$-\frac{12}{k} + 6 = 4$$
  4. Solve for $k$: $$-\frac{12}{k} = 4 - 6 = -2 \implies 2k = 12 \implies k = 6$$

Final Answer: $k = 6$

Exercise 9 Paper 1 Trigonometric Bounded Area

Find the exact area enclosed between the curves $y = \cos x$ and $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{4}$.

View Complete Solution
  1. Identify curve positions on $\left[0, \frac{\pi}{4}\right]$: $$\text{At } x = 0: \cos(0) = 1, \sin(0) = 0 \implies \cos x \ge \sin x \text{ on } \left[0, \frac{\pi}{4}\right]$$
  2. Set up the area integral: $$\text{Area} = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx$$
  3. Integrate each term: $$= [\sin x - (-\cos x)]_{0}^{\pi/4} = [\sin x + \cos x]_{0}^{\pi/4}$$
  4. Evaluate at limits: $$\text{Upper limit } \left(x = \frac{\pi}{4}\right): \quad \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$$ $$\text{Lower limit } (x = 0): \quad \sin(0) + \cos(0) = 0 + 1 = 1$$ $$\text{Area} = \sqrt{2} - 1$$

Final Answer: $\sqrt{2} - 1$ square units ($\approx 0.414$)

Exercise 10 Paper 1 / Paper 2 Multi-part Cubic Analysis

Consider the cubic polynomial function $f(x) = x(x - 2)(x - 4) = x^3 - 6x^2 + 8x$.

(a) Write down the $x$-intercepts of the curve.

(b) Show by integration that $\int_{0}^{4} f(x) \, dx = 0$.

(c) Interpret geometrically why the integral in (b) is zero.

(d) Calculate the total geometric area enclosed between the curve and the $x$-axis from $x = 0$ to $x = 4$.

View Complete Solution
  1. Part (a): Setting $f(x) = 0 \implies x = 0, \quad x = 2, \quad x = 4$.
  2. Part (b): $$\int_{0}^{4} (x^3 - 6x^2 + 8x) \, dx = \left[ \frac{x^4}{4} - 2x^3 + 4x^2 \right]_{0}^{4}$$ $$= \left( \frac{4^4}{4} - 2(4^3) + 4(4^2) \right) - 0 = (64 - 128 + 64) - 0 = 0 \quad \text{[Shown]}$$
  3. Part (c): The region from $x = 0$ to $x = 2$ lies above the $x$-axis (positive signed area $+A$), and the region from $x = 2$ to $x = 4$ is congruent and lies below the $x$-axis (negative signed area $-A$). In the net integral, these two signed areas cancel each other out completely: $+A + (-A) = 0$.
  4. Part (d): Evaluate the positive region $[0, 2]$: $$A_1 = \int_{0}^{2} (x^3 - 6x^2 + 8x) \, dx = \left[ \frac{2^4}{4} - 2(2^3) + 4(2^2) \right] = 4 - 16 + 16 = 4$$ By symmetry, the second region has geometric area $A_2 = |-4| = 4$. $$\text{Total Area} = A_1 + A_2 = 4 + 4 = 8$$

Final Answer: (a) $x = 0, 2, 4$; (b) [Shown]; (c) Congruent regions above and below axis cancel; (d) $8$ square units

Formula Booklet SL 5.10

Standard Indefinite Integrals

  • Power Rule: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$
  • Reciprocal / Natural Log: $$\int \frac{1}{x} \, dx = \ln|x| + C \quad (x > 0)$$
  • Exponential: $$\int e^x \, dx = e^x + C$$
  • Trigonometric: $$\int \cos x \, dx = \sin x + C, \qquad \int \sin x \, dx = -\cos x + C$$
Must Memorize (Inspection)

Linear Composite Form Rules

  • General Linear Rule: $$\int f(ax + b) \, dx = \frac{1}{a} F(ax + b) + C$$
  • Linear Power: $$\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C \quad (n \neq -1)$$
  • Linear Exponential: $$\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C$$
  • Linear Denominator: $$\int \frac{1}{ax + b} \, dx = \frac{1}{a}\ln|ax + b| + C$$
  • Linear Trigonometric: $$\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + C, \quad \int \sin(kx) \, dx = -\frac{1}{k}\cos(kx) + C$$
Formula Booklet SL 5.5, 5.11

Definite Integrals & Area Enclosures

  • Fundamental Theorem: $$\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)$$
  • Total Geometric Area: $$\text{Area} = \int_a^b |f(x)| \, dx = \sum \left| \int_{x_i}^{x_{i+1}} f(x) \, dx \right|$$
  • Area Enclosed Between Two Curves: $$\text{Area} = \int_a^b [f(x) - g(x)] \, dx \quad \text{where } f(x) \ge g(x)$$
  • Integral Properties: $$\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx, \qquad \int_a^c f(x) \, dx + \int_c^b f(x) \, dx = \int_a^b f(x) \, dx$$

Critical IB Examination Pitfalls in Integration

  • Forgetting $+C$ on Indefinite Integrals: IB markschemes penalize the missing arbitrary constant with an accuracy penalty ($A0$) on Paper 1.
  • The Negative Sign on $\int \sin x \, dx$: Differentiating $\sin x$ gives $+\cos x$, but integrating $\sin x$ gives $-\cos x + C$. Do not confuse the two operations!
  • Omitting the $\frac{1}{a}$ Factor: When evaluating $\int e^{3x} dx$, writing $3e^{3x}$ instead of $\frac{1}{3}e^{3x}$ is one of the most common student errors in Paper 1.
  • Ignoring Negative Regions in Area Questions: A direct evaluation of $\int_a^b f(x) dx$ gives the net signed area, NOT the geometric area. You must solve for intercepts $f(x) = 0$ and split intervals where the graph dips below the $x$-axis.
  • Forgetting to Change Limits in Substitution: If substituting $u = g(x)$, you must convert $x$-limits to $u$-limits ($u_a = g(a)$). Never evaluate an antiderivative in $u$ using the original $x$-limits!

Section A: Diagnostic Check

4 core concept check questions. Immediate feedback and scoring.

Score: 0 / 4

Question 1

What is the antiderivative of $f(x) = 6x^2 - 4x + \frac{3}{x^2}$?

Correct! Rewriting $\frac{3}{x^2}$ as $3x^{-2}$ yields $\int 3x^{-2} dx = \frac{3x^{-1}}{-1} = -\frac{3}{x}$. Thus $\int f(x) dx = 2x^3 - 2x^2 - \frac{3}{x} + C$.

Question 2

In evaluating $\int_{1}^{3} 2x(x^2 - 1)^4 \, dx$ using the substitution $u = x^2 - 1$, what is the transformed integral in terms of $u$?

Correct! $du = 2x dx$. The lower limit becomes $u(1) = 1^2 - 1 = 0$, and the upper limit becomes $u(3) = 3^2 - 1 = 8$, giving $\int_0^8 u^4 du$.

Question 3

If $\int_{0}^{2} f(x) \, dx = 5$ and $\int_{2}^{5} f(x) \, dx = -3$, and it is given that $f(x) \le 0$ on $[2, 5]$, what is the total geometric area enclosed by $y = f(x)$ and the $x$-axis between $x = 0$ and $x = 5$?

Correct! The net integral is $5 + (-3) = 2$, but total geometric area takes the absolute value of the region below the axis: $\text{Area} = 5 + |-3| = 5 + 3 = 8$.

Question 4

The area enclosed between $y = x^2$ and $y = 2x$ is given by:

Correct! The intersection points are $x = 0$ and $x = 2$. On this interval, the line $y = 2x$ is above the parabola $y = x^2$, so $\text{Area} = \int_0^2 (2x - x^2) dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

The derivative of a function is given by $f'(x) = 3x^2 - \frac{8}{x^3}$ for $x \neq 0$.

(a) Find an expression for $f(x)$, given that the graph of $f$ passes through the point $(1, 6)$. [4 marks]
(b) Find the equation of the tangent line to the graph of $f$ at $x = 1$. [2 marks]
View Markscheme
(a)
Rewrite $f'(x) = 3x^2 - 8x^{-3}$ M1
Integrate: $f(x) = x^3 - 8\left(\frac{x^{-2}}{-2}\right) + C = x^3 + \frac{4}{x^2} + C$ A1
Substitute point $(1, 6)$: $6 = 1^3 + \frac{4}{1^2} + C \implies 6 = 5 + C \implies C = 1$ M1
$$\mathbf{f(x) = x^3 + \frac{4}{x^2} + 1}$$ A1

(b)
Evaluate gradient at $x = 1$: $m = f'(1) = 3(1)^2 - 8(1)^{-3} = 3 - 8 = -5$ M1
Using point $(1, 6)$: $y - 6 = -5(x - 1) \implies \mathbf{y = -5x + 11} \quad (\text{or } 5x + y - 11 = 0)$ A1
Paper 2 (Calculator Allowed) [7 marks]

Let $f(x) = 4 - x^2$ and $g(x) = 2 - x$.

(a) Find the coordinates of the two points of intersection of the graphs of $f$ and $g$. [3 marks]
(b) Calculate the exact area of the finite region bounded by the two graphs. [4 marks]
View Markscheme
(a)
Set $4 - x^2 = 2 - x \implies x^2 - x - 2 = 0$ M1
$(x - 2)(x + 1) = 0 \implies x = -1 \quad \text{and} \quad x = 2$ A1
Find corresponding $y$-values: $g(-1) = 2 - (-1) = 3$ and $g(2) = 2 - 2 = 0$
Coordinates are $\mathbf{(-1, 3)} \quad \text{and} \quad \mathbf{(2, 0)}$ A1

(b)
Set up area integral: $\text{Area} = \int_{-1}^{2} [(4 - x^2) - (2 - x)] \, dx$ M1
$= \int_{-1}^{2} (2 + x - x^2) \, dx$ A1
$= \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}$ M1
$=\left( 4 + 2 - \frac{8}{3} \right) - \left( -2 + \frac{1}{2} + \frac{1}{3} \right) = \frac{10}{3} - \left( -\frac{7}{6} \right) = \mathbf{\frac{27}{6} = \frac{9}{2} = 4.5\text{ square units}}$ A1

Unit Overview

All Lessons in Calculus (Topic 5)

5.1, 5.3 Limits & First Derivatives Derivative as limiting rate of change, power rule, and secant geometry. 5.2, 5.7, 5.8 Increasing, Decreasing & Stationary Points Monotonicity tests, second derivatives, local extrema, and inflection points. 5.4, 5.6 Tangents, Normals & Rules Product, quotient, and chain rules with tangent and normal line equations. 5.9 Kinematics with Derivatives Motion in a line: displacement, velocity, acceleration, and total distance. 5.5, 5.10, 5.11 Integration & Area Anti-differentiation, definite integrals, and area enclosed between curves.