Integration (Calculus).

Grade 12 Mathematics: Section 1.8 - Integration (Finding Area Under Curves; Reversing Differentiation)

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Lesson Objectives

  • Understand integration as the reverse process of differentiation
  • Apply the power rule for integration
  • Understand the constant of integration and its role
  • Evaluate definite integrals using limits of integration
  • Find the area under a curve using definite integration
  • Calculate areas between curves
  • Apply integration to solve real-world problems involving accumulation

Introduction to Integration

Integration is the reverse process of differentiation. While differentiation finds the rate of change of a function, integration finds the original function from its derivative. Integration is also used to calculate the area under a curve, total accumulated change, volumes of revolution, and many other quantities in mathematics and science.

The Fundamental Relationship
Differentiation: $\frac{d}{dx}f(x) = f'(x)$
Integration: $\int f'(x) \ dx = f(x) + C$
Where $C$ is the constant of integration.
Key Definitions:
• Integration: The process of finding the antiderivative of a function.
• Indefinite Integral: $\int f(x) \ dx = F(x) + C$, where $F'(x) = f(x)$.
• Definite Integral: $\int_a^b f(x) \ dx = F(b) - F(a)$, which gives the area under $f(x)$ from $x=a$ to $x=b$.
• Integrand: The function $f(x)$ being integrated.
• Constant of Integration ($C$): An arbitrary constant added because the derivative of a constant is zero.

Indefinite Integration (Reversing Differentiation)

Since differentiation finds the derivative, indefinite integration (antidifferentiation) finds the original function. Because the derivative of any constant is zero, we must always add a constant $C$ to the result.

Step-by-Step Method for Indefinite Integration:
1. Increase the power of $x$ by 1.
2. Divide by the new power.
3. Add the constant of integration $C$.
Formula: $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$
Example 1: Basic Power Rule Integration
Problem: Find $\int x^3 \ dx$.

Formula: $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C$

Solution:
$\int x^3 \ dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$
Answer: $\frac{x^4}{4} + C$
Example 2: Integrating a Constant
Problem: Find $\int 5 \ dx$.

Formula: $\int k \ dx = kx + C$

Solution:
$\int 5 \ dx = 5x + C$
Answer: $5x + C$
Example 3: Integrating a Negative Power
Problem: Find $\int x^{-2} \ dx, \ x \neq 0$.

Solution:
$\int x^{-2} \ dx = \frac{x^{-2+1}}{-2+1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$
Answer: $-\frac{1}{x} + C$
Example 4: Integrating a Function (Reversing Differentiation)
Problem: If $f'(x) = 6x^2 - 4x + 3$, find $f(x)$.

Solution:
$f(x) = \int (6x^2 - 4x + 3) \ dx = 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 3x + C = 2x^3 - 2x^2 + 3x + C$
Answer: $f(x) = 2x^3 - 2x^2 + 3x + C$
Watch Out!
The power rule for integration does not work for $n = -1$ (i.e., $\int x^{-1} dx = \int \frac{1}{x} dx = \ln|x| + C$). This will be covered in more advanced topics.

Practice for Indefinite Integration

  1. Find $\int x^5 \ dx$.
  2. Find $\int 7 \ dx$.
  3. Find $\int (3x^2 + 2x - 5) \ dx$.
  4. Find $\int x^{-3} \ dx$.
  5. If $f'(x) = 8x^3 - 6x$, find $f(x)$.

Rules of Integration

Just as with differentiation, there are rules that make integration easier. These rules are essentially the reverse of differentiation rules.

Basic Integration Rules
1. Power Rule: $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
2. Constant Multiple Rule: $\int k f(x) \ dx = k \int f(x) \ dx$
3. Sum/Difference Rule: $\int [f(x) \pm g(x)] \ dx = \int f(x) \ dx \pm \int g(x) \ dx$
4. Constant Rule: $\int k \ dx = kx + C$
Example 1: Constant Multiple Rule
Problem: Find $\int 4x^3 \ dx$.

Formula: $\int k f(x) \ dx = k \int f(x) \ dx$

Solution:
$\int 4x^3 \ dx = 4 \int x^3 \ dx = 4 \cdot \frac{x^4}{4} + C = x^4 + C$
Answer: $x^4 + C$
Example 2: Sum Rule
Problem: Find $\int (5x^4 + 3x^2) \ dx$.

Formula: $\int [f(x) + g(x)] \ dx = \int f(x) \ dx + \int g(x) \ dx$

Solution:
$\int 5x^4 \ dx = 5 \cdot \frac{x^5}{5} = x^5$, $\int 3x^2 \ dx = 3 \cdot \frac{x^3}{3} = x^3$
Answer: $x^5 + x^3 + C$
Example 3: Difference Rule
Problem: Find $\int (6x^5 - 2x) \ dx$.

Solution:
$\int 6x^5 \ dx = 6 \cdot \frac{x^6}{6} = x^6$, $\int 2x \ dx = 2 \cdot \frac{x^2}{2} = x^2$
Answer: $x^6 - x^2 + C$
Example 4: Polynomial Integration
Problem: Find $\int (2x^5 - 4x^3 + 7x - 9) \ dx$.

Solution:
$\int 2x^5 \ dx = 2 \cdot \frac{x^6}{6} = \frac{x^6}{3}$
$\int -4x^3 \ dx = -4 \cdot \frac{x^4}{4} = -x^4$
$\int 7x \ dx = 7 \cdot \frac{x^2}{2} = \frac{7x^2}{2}$
$\int -9 \ dx = -9x$
Answer: $\frac{x^6}{3} - x^4 + \frac{7x^2}{2} - 9x + C$

Practice for Integration Rules

  1. Find $\int 8x^6 \ dx$.
  2. Find $\int (4x^3 - 6x^2 + 2) \ dx$.
  3. Find $\int (9x^8 + 5) \ dx$.
  4. Find $\int (x^4 - 3x^2 + 4x) \ dx$.
  5. Find $\int (2x^{-3} + 5x^{-1}?)$ but note: for $x^{-1}$ use special rule $\ln|x|$ - just do the $x^{-3}$ part: $\int 2x^{-3} \ dx$.

The Constant of Integration

The constant of integration $C$ is essential because the derivative of any constant is zero. When we find an antiderivative, there are infinitely many possibilities differing by a constant. To determine a specific function, we need additional information (like a point on the original curve).

Finding a Particular Antiderivative:
1. Integrate to find $F(x) + C$.
2. Use the given condition (e.g., $F(a) = b$) to solve for $C$.
3. Substitute $C$ back to get the particular antiderivative.
Example 1: Finding a Particular Function
Problem: Find $f(x)$ if $f'(x) = 4x^3$ and $f(1) = 5$.

Step 1: $f(x) = \int 4x^3 \ dx = x^4 + C$
Step 2: $f(1) = 1^4 + C = 1 + C = 5$ → $C = 4$
Answer: $f(x) = x^4 + 4$
Example 2: Finding a Function from Gradient
Problem: The gradient of a curve is given by $\frac{dy}{dx} = 6x - 2$. The curve passes through the point $(2, 7)$. Find the equation of the curve.

Step 1: $y = \int (6x - 2) \ dx = 3x^2 - 2x + C$
Step 2: When $x = 2$, $y = 7$: $3(4) - 2(2) + C = 12 - 4 + C = 8 + C = 7$ → $C = -1$
Answer: $y = 3x^2 - 2x - 1$

Practice for The Constant of Integration

  1. Find $f(x)$ if $f'(x) = 2x$ and $f(0) = 3$.
  2. Find $f(x)$ if $f'(x) = 3x^2 - 4$ and $f(1) = 2$.
  3. A curve has gradient $\frac{dy}{dx} = 4x + 1$ and passes through $(0, 5)$. Find its equation.
  4. Find $f(x)$ if $f'(x) = x^2 - 2x$ and $f(2) = 4$.
  5. A particle's velocity is $v(t) = 6t^2 - 2t$. If its initial position is $s(0) = 3$, find $s(t)$.

Definite Integration

A definite integral has limits of integration and gives a numerical value. It represents the net area under the curve between $x = a$ and $x = b$. The formula is $\int_a^b f(x) \ dx = F(b) - F(a)$, where $F(x)$ is an antiderivative.

Step-by-Step Method for Definite Integrals:
1. Find the indefinite integral $F(x) + C$ (the constant cancels out).
2. Evaluate $F(b) - F(a)$.
3. Simplify to get the numerical value.
Example 1: Basic Definite Integral
Problem: Evaluate $\int_1^3 x^2 \ dx$.

Step 1: $\int x^2 \ dx = \frac{x^3}{3}$
Step 2: $\left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$
Answer: $\frac{26}{3}$
Example 2: Definite Integral of a Polynomial
Problem: Evaluate $\int_0^2 (3x^2 - 2x + 1) \ dx$.

Step 1: $\int (3x^2 - 2x + 1) \ dx = x^3 - x^2 + x$
Step 2: $[x^3 - x^2 + x]_0^2 = (8 - 4 + 2) - (0 - 0 + 0) = 6 - 0 = 6$
Answer: $6$
Example 3: Definite Integral with Negative Limits
Problem: Evaluate $\int_{-1}^1 (4x^3) \ dx$.

Step 1: $\int 4x^3 \ dx = x^4$
Step 2: $[x^4]_{-1}^1 = 1^4 - (-1)^4 = 1 - 1 = 0$
Answer: $0$ (odd function over symmetric interval)
Watch Out!
For definite integrals, the constant of integration cancels out. Do not include $+C$ when evaluating $F(b) - F(a)$. Always substitute the limits in the correct order: $F(b) - F(a)$.

Practice for Definite Integration

  1. Evaluate $\int_1^4 x^3 \ dx$.
  2. Evaluate $\int_0^3 (2x^2 + 1) \ dx$.
  3. Evaluate $\int_{-2}^2 (6x^2) \ dx$.
  4. Evaluate $\int_1^5 (4x - 3) \ dx$.
  5. Evaluate $\int_0^4 (x^2 - 3x + 2) \ dx$.

Finding Area Under a Curve

The definite integral $\int_a^b f(x) \ dx$ gives the net area between the curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$. Area above the x-axis is positive, and area below is negative. To find total area, split the integral at x-intercepts.

Finding Area Under a Curve:
1. Identify the limits $a$ and $b$ (where the curve meets the x-axis if finding total area).
2. Set up the definite integral $\int_a^b f(x) \ dx$.
3. Evaluate the integral.
4. If $f(x)$ is negative on the interval, the integral gives a negative value; take the absolute value for area.
Example 1: Area Under a Curve (Above x-axis)
Problem: Find the area under $y = x^2$ from $x = 0$ to $x = 2$.

Formula: $A = \int_0^2 x^2 \ dx$

Solution:
$\int_0^2 x^2 \ dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$ square units
Answer: $\frac{8}{3}$ square units
Example 2: Area Below the x-axis
Problem: Find the area between $y = x - 2$ and the x-axis from $x = 0$ to $x = 2$.

Step 1: The curve is below the x-axis on $[0,2]$.
Step 2: $\int_0^2 (x - 2) \ dx = \left[ \frac{x^2}{2} - 2x \right]_0^2 = (2 - 4) - (0 - 0) = -2$
Step 3: Area = $|-2| = 2$ square units
Answer: 2 square units
Example 3: Total Area (Above and Below)
Problem: Find the total area between $y = x^2 - 4$ and the x-axis from $x = 0$ to $x = 3$.

Step 1: Find x-intercepts: $x^2 - 4 = 0$ → $x = \pm 2$. On $[0,3]$, the curve crosses at $x = 2$.
Step 2: Split the integral: $\int_0^2 (x^2 - 4) \ dx$ (negative) and $\int_2^3 (x^2 - 4) \ dx$ (positive).
Step 3: $\int_0^2 (x^2 - 4) \ dx = \left[ \frac{x^3}{3} - 4x \right]_0^2 = (\frac{8}{3} - 8) - 0 = \frac{8}{3} - \frac{24}{3} = -\frac{16}{3}$
$\int_2^3 (x^2 - 4) \ dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = (9 - 12) - (\frac{8}{3} - 8) = (-3) - (\frac{8}{3} - \frac{24}{3}) = -3 - (-\frac{16}{3}) = -3 + \frac{16}{3} = -\frac{9}{3} + \frac{16}{3} = \frac{7}{3}$
Step 4: Total area = $\frac{16}{3} + \frac{7}{3} = \frac{23}{3}$ square units
Answer: $\frac{23}{3}$ square units
Example 4: Area Between Two Curves
Problem: Find the area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$.

Formula: $A = \int_a^b (y_{\text{top}} - y_{\text{bottom}}) \ dx$
Solution:
On $[0,1]$, $x \geq x^2$, so top curve is $y = x$, bottom is $y = x^2$.
$A = \int_0^1 (x - x^2) \ dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$
Answer: $\frac{1}{6}$ square units
Watch Out!
The definite integral gives net area (area above minus area below). To find total area, split the integral at points where the function crosses the x-axis.
Area Under Curve Summary
SituationMethodExample
Curve above x-axis ($f(x) \geq 0$)$A = \int_a^b f(x) \ dx$$y = x^2$ on $[0,2]$
Curve below x-axis ($f(x) \leq 0$)$A = \left| \int_a^b f(x) \ dx \right|$$y = x-2$ on $[0,2]$
Curve crosses x-axisSplit at x-intercepts; add absolute values$y = x^2 - 4$ on $[0,3]$
Between two curves$A = \int_a^b (y_{\text{top}} - y_{\text{bottom}}) \ dx$$y = x$ and $y = x^2$

Practice for Area Under a Curve

  1. Find the area under $y = x^3$ from $x = 0$ to $x = 2$.
  2. Find the area under $y = 4 - x^2$ from $x = -2$ to $x = 2$.
  3. Find the total area between $y = x^2 - 1$ and the x-axis from $x = -2$ to $x = 2$.
  4. Find the area between $y = x^2$ and $y = 2x$ from $x = 0$ to $x = 2$.
  5. Find the area between $y = x + 1$ and $y = x^2 - 1$ from $x = -1$ to $x = 2$.

Integrating (ax + b)^n (Reverse Chain Rule)

When integrating a function of the form $(ax + b)^n$, we use the reverse of the chain rule.

Formula: $\int (ax + b)^n \ dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C, \quad n \neq -1$
Example 1: Linear Function Squared
Problem: Find $\int (2x + 1)^3 \ dx$.
Formula: $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$
Solution: Here $a = 2$, $n = 3$. $\int (2x+1)^3 dx = \frac{(2x+1)^{4}}{2(4)} + C = \frac{(2x+1)^4}{8} + C$.
Example 2: Function in Denominator
Problem: Find $\int \frac{1}{(3x-2)^2} \ dx = \int (3x-2)^{-2} dx$.
Solution: $n = -2$, $a = 3$. $\int (3x-2)^{-2} dx = \frac{(3x-2)^{-1}}{3(-1)} + C = -\frac{1}{3(3x-2)} + C$.
Remember to divide by the derivative of the inner function (a) and adjust the power correctly.

Practice for (ax + b)^n

  1. Find $\int (4x + 3)^2 \ dx$.
  2. Find $\int \frac{1}{(x+5)^3} \ dx$.

Integration of Trigonometric Functions

Since differentiation and integration are reverses, we can derive the integrals of standard trig functions.

Standard Trigonometric Integrals
$\int \sin x \ dx = -\cos x + C$
$\int \cos x \ dx = \sin x + C$
$\int \sec^2 x \ dx = \tan x + C$
$\int \csc^2 x \ dx = -\cot x + C$
$\int \sec x \tan x \ dx = \sec x + C$
$\int \csc x \cot x \ dx = -\csc x + C$
Example 1: Basic Trig Integration
Problem: Find $\int (3\sin x + 4\cos x) \ dx$.
Solution: $3(-\cos x) + 4(\sin x) + C = -3\cos x + 4\sin x + C$.
Example 2: Integration using Reverse Chain Rule with Trig
Problem: Find $\int \cos(2x + \pi) \ dx$.
Formula: $\int \cos(ax+b) dx = \frac{\sin(ax+b)}{a} + C$
Solution: $\frac{\sin(2x + \pi)}{2} + C$.
Be careful with signs: The derivative of $\cos x$ is $-\sin x$, so the integral of $\sin x$ introduces a negative sign.

Practice for Trig Integration

  1. Find $\int (2\sin x - \cos x) \ dx$.
  2. Find $\int \sin(3x) \ dx$.
  3. Find $\int \sec^2 (4x) \ dx$.

Integration by Parts

Integration by parts is used for products of functions. It is derived from the product rule for differentiation.

Formula: $\int u \ dv = uv - \int v \ du$
Choose $u$ as the function that becomes simpler when differentiated, and $dv$ as the part that can be integrated easily.
Example 1: $x \sin x$
Problem: Find $\int x \sin x \ dx$.
Step 1: Choose $u = x$ (so $du = dx$) and $dv = \sin x \ dx$ (so $v = -\cos x$).
Step 2: Apply formula: $\int u \ dv = uv - \int v \ du = x(-\cos x) - \int (-\cos x) dx = -x\cos x + \int \cos x \ dx$.
Step 3: $\int \cos x \ dx = \sin x + C$.
Answer: $-x\cos x + \sin x + C$.
Example 2: $x e^{2x}$
Problem: Find $\int x e^{2x} \ dx$.
Step 1: Choose $u = x$ ($du = dx$) and $dv = e^{2x}dx$ ($v = \frac{e^{2x}}{2}$).
Step 2: $\int u \ dv = x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} dx = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + C$.
Answer: $\frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}}{4}(2x - 1) + C$.
The traditional choice for $u$ is LIATE: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Choose $u$ as the function that comes first in this list.

Practice for Integration by Parts

  1. Find $\int x \cos x \ dx$.
  2. Find $\int x^2 e^x \ dx$.
  3. Find $\int \ln x \ dx$. (Hint: Write $\ln x$ as $1 \cdot \ln x$).

Methods & Techniques

Verification / Checking Strategy:
1. For indefinite integrals: Differentiate your answer to check if you get back the original integrand.
2. For definite integrals: Check that the value is reasonable (e.g., area should be positive).
3. For area problems: Sketch the curve to see if the integral gives net or total area.
4. For particular solutions: Substitute the given point to verify $C$.
Example: Checking an Indefinite Integral
Original: $\int 3x^2 \ dx = x^3 + C$

Check: Differentiate $x^3 + C$ → $3x^2$ ✓
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Forgetting the constant of integration $C$ in indefinite integrals → Solution: Always add $+C$.
• Pitfall 2: Incorrectly applying the power rule (dividing by $n+1$, not $n$) → Solution: $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C$.
• Pitfall 3: Mixing up limits in definite integrals → Solution: Always do $F(b) - F(a)$.
• Pitfall 4: Taking the integral of a negative region as negative area → Solution: Take absolute value for total area or split at intercepts.
• Pitfall 5: Forgetting that $\int x^{-1} dx = \ln|x| + C$ (not covered here but note for extension).

Technique Practice

  1. Verify that $\int 4x^3 \ dx = x^4 + C$ by differentiation.
  2. Check the definite integral $\int_1^3 x^2 \ dx = \frac{26}{3}$.
  3. Identify the error: A student integrated $\int (x^2 + 2x) \ dx$ as $\frac{x^3}{3} + x^2$. Correct it.
  4. For area under $y = x^2 - 4$ from $0$ to $3$, why do we need to split at $x = 2$?

Real-World Applications

Application 1: Physics - Displacement from Velocity
Scenario: A particle moves with velocity $v(t) = 3t^2 - 2t$ m/s. Find the displacement between $t = 1$ and $t = 3$.
Problem: Displacement = $\int_1^3 v(t) \ dt$.

Solution:
$\int_1^3 (3t^2 - 2t) \ dt = \left[ t^3 - t^2 \right]_1^3 = (27 - 9) - (1 - 1) = 18 - 0 = 18$ m
Answer: 18 metres
Application 2: Economics - Consumer Surplus
Scenario: The demand function is $p = 100 - 2q$. Find the consumer surplus at $q = 30$.
Problem: Consumer surplus = $\int_0^{q_0} (D(q) - p_0) \ dq$.

Solution:
At $q = 30$, $p_0 = 100 - 60 = 40$
CS = $\int_0^{30} (100 - 2q - 40) \ dq = \int_0^{30} (60 - 2q) \ dq = \left[ 60q - q^2 \right]_0^{30} = 1800 - 900 = 900$
Answer: 900
Application 3: Volume of a Solid (Simple Case)
Scenario: A solid has cross-sectional area $A(x) = \pi x^2$. Find the volume from $x = 0$ to $x = 3$.
Problem: Volume = $\int_0^3 A(x) \ dx$.

Solution:
$V = \int_0^3 \pi x^2 \ dx = \pi \left[ \frac{x^3}{3} \right]_0^3 = \pi \left( \frac{27}{3} - 0 \right) = 9\pi$
Answer: $9\pi$ cubic units
Application 4: Average Value of a Function
Scenario: Find the average value of $f(x) = x^2$ from $x = 0$ to $x = 3$.
Problem: Average value = $\frac{1}{b-a} \int_a^b f(x) \ dx$.

Solution:
$\frac{1}{3-0} \int_0^3 x^2 \ dx = \frac{1}{3} \cdot \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3$
Answer: 3

Cross-Curricular Connections

  • Physics: Displacement, work, centre of mass, moment of inertia
  • Economics: Consumer surplus, producer surplus, total revenue
  • Biology: Population growth accumulation, drug concentration over time
  • Engineering: Area of cross-sections, volume of revolution, fluid flow

Cumulative Practice Exercises

  1. Find $\int x^4 \ dx$.
  2. Find $\int (4x^3 - 2x + 5) \ dx$.
  3. Find $f(x)$ if $f'(x) = 6x - 4$ and $f(2) = 5$.
  4. Evaluate $\int_1^4 x^2 \ dx$.
  5. Evaluate $\int_0^2 (3x^2 - 4x + 1) \ dx$.
  6. Find the area under $y = x^3$ from $x = 0$ to $x = 2$.
  7. Find the total area between $y = x^2 - 4$ and the x-axis from $x = -3$ to $x = 3$.
  8. Find the area between $y = x^2$ and $y = 2x$ from $x = 0$ to $x = 2$.
  9. A particle's velocity is $v(t) = 4t - t^2$. Find the displacement from $t = 0$ to $t = 4$.
  10. Find the average value of $f(x) = 2x$ from $x = 1$ to $x = 5$.
  11. Find $f(x)$ if $f'(x) = 8x^3 - 6x$ and $f(0) = 2$.
  12. Evaluate $\int_{-1}^2 (x^2 + 2x) \ dx$.
  13. Error analysis: A student integrated $\int (4x^3 + 2) \ dx$ as $x^4 + 2x + C$. Is this correct? If not, correct it.
  14. Find the area between $y = x + 2$ and $y = x^2$ from $x = -1$ to $x = 2$.
  15. A curve has gradient $\frac{dy}{dx} = 2x - 3$ and passes through $(1, 4)$. Find its equation.
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Answers to Cumulative Exercises

  1. Answer: $\frac{x^5}{5} + C$
  2. Answer: $x^4 - x^2 + 5x + C$
  3. Answer: $f(x) = 3x^2 - 4x + C$, $f(2) = 12 - 8 + C = 4 + C = 5$ → $C = 1$, so $f(x) = 3x^2 - 4x + 1$
  4. Answer: $\left[ \frac{x^3}{3} \right]_1^4 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21$
  5. Answer: $\left[ x^3 - 2x^2 + x \right]_0^2 = (8 - 8 + 2) - 0 = 2$
  6. Answer: $\int_0^2 x^3 \ dx = \left[ \frac{x^4}{4} \right]_0^2 = \frac{16}{4} - 0 = 4$ square units
  7. Answer: $x^2 - 4 = 0$ → $x = \pm 2$. Split into three intervals: $[-3,-2]$, $[-2,2]$, $[2,3]$. Due to symmetry: $2 \times \int_2^3 (x^2-4) dx$? Actually total area = $\int_{-3}^{-2} (x^2-4) dx + \int_{-2}^2 (4-x^2) dx + \int_2^3 (x^2-4) dx$. Compute: $\int_2^3 (x^2-4) dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = (9-12) - (\frac{8}{3}-8) = -3 - (\frac{8}{3}-\frac{24}{3}) = -3 - (-\frac{16}{3}) = -3 + \frac{16}{3} = \frac{7}{3}$. The area from -3 to -2 is the same = $\frac{7}{3}$. And $\int_{-2}^2 (4-x^2) dx = 2\int_0^2 (4-x^2) dx = 2\left[ 4x - \frac{x^3}{3} \right]_0^2 = 2(8 - \frac{8}{3}) = 2(\frac{16}{3}) = \frac{32}{3}$. Total = $\frac{7}{3} + \frac{32}{3} + \frac{7}{3} = \frac{46}{3}$ square units.
  8. Answer: $\int_0^2 (2x - x^2) \ dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$ square units
  9. Answer: $\int_0^4 (4t - t^2) \ dt = \left[ 2t^2 - \frac{t^3}{3} \right]_0^4 = 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$ m
  10. Answer: $\frac{1}{5-1} \int_1^5 2x \ dx = \frac{1}{4} \left[ x^2 \right]_1^5 = \frac{1}{4}(25 - 1) = \frac{24}{4} = 6$
  11. Answer: $f(x) = 2x^4 - 3x^2 + 2$
  12. Answer: $\left[ \frac{x^3}{3} + x^2 \right]_{-1}^2 = (\frac{8}{3} + 4) - (-\frac{1}{3} + 1) = (\frac{8}{3} + \frac{12}{3}) - (-\frac{1}{3} + \frac{3}{3}) = \frac{20}{3} - \frac{2}{3} = \frac{18}{3} = 6$
  13. Answer: The student is correct! $\int 4x^3 \ dx = x^4$, $\int 2 \ dx = 2x$, so $x^4 + 2x + C$ is correct.
  14. Answer: Find intersection points: $x + 2 = x^2$ → $x^2 - x - 2 = 0$ → $(x-2)(x+1)=0$ → $x = -1, 2$. On $[-1,2]$, $x+2 \geq x^2$. Area = $\int_{-1}^2 (x+2 - x^2) dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2 = (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = (6 - \frac{8}{3}) - (-\frac{6}{6} + \frac{?}{.}$ Let's compute carefully: First term: $2 + 4 = 6$, $6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$. Second term: $\frac{1}{2} - 2 = -\frac{3}{2}$, $-\frac{3}{2} + \frac{1}{3} = -\frac{9}{6} + \frac{2}{6} = -\frac{7}{6}$. So $(\frac{10}{3}) - (-\frac{7}{6}) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$ square units.
  15. Answer: $y = \int (2x - 3) \ dx = x^2 - 3x + C$. When $x = 1$, $y = 4$: $1 - 3 + C = -2 + C = 4$ → $C = 6$. So $y = x^2 - 3x + 6$.

Conclusion & Summary

Integration is the reverse process of differentiation. Indefinite integration finds the family of functions that differentiate to the given function, always including a constant $+C$. Definite integration gives a numerical value representing the net area under a curve between two limits. The area under a curve can be found using definite integrals, and for regions below the x-axis or between curves, appropriate adjustments are needed.

Key Takeaways:
1. Indefinite Integral: $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$)
2. Constant of Integration $C$: Added because the derivative of a constant is zero.
3. Definite Integral: $\int_a^b f(x) \ dx = F(b) - F(a)$
4. Area Under Curve: $A = \int_a^b f(x) \ dx$ (if $f(x) \geq 0$ on $[a,b]$)
5. Total Area: Split integral at x-intercepts where $f(x)$ changes sign.
6. Area Between Curves: $A = \int_a^b (y_{\text{top}} - y_{\text{bottom}}) \ dx$
7. Applications: Displacement, work, consumer surplus, average value.

Keep practising integration. It is the foundation for much of advanced mathematics, physics, and engineering!

Video Resource

Watch this video for more examples of integration, definite integrals, and finding area under curves.

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