Integrals — Class 12 Maths Notes

Topics covered in Integrals

1. 1.Introduction of Integrals

   Introduction to Integrals

   Integrals are a fundamental concept in mathematics, especially in calculus. They are used to find many things, such as the area under a curve, the total accumulation of quantities, and in many other applications in science and engineering. Let's explore this concept starting with a simple example from daily life.

   Imagine you're trying to figure out how much paint you need to cover a strangely shaped wall. By breaking the wall down into very small sections, each section resembling a rectangle, and then adding up all these little "slices" of paint, you can find out the total amount of paint needed. This process is similar to integration, where you sum up infinitesimally small slices to find the whole.

   How It Works in Mathematics

   In mathematics, integration is essentially the process of adding up parts to find the whole. This is often visualized as finding the area under a curve on a graph. When you integrate a function, you are finding the sum of an infinite number of infinitesimally small areas under the curve of that function.

   Real-life Applications

   1. Physics: Calculating distances traveled over time when speeds change.
   2. Economics: Determining total cost when given a cost function.
   3. Biology: Modeling population growths over time.
   4. Engineering: Analyzing forces acting on a structure.

   Careers and Industries

   Understanding integrals is crucial in fields like:

   • Engineering: For designing and analyzing systems and structures.
   • Economics: For modeling economic growth and optimizing resources.
   • Environmental Science: For modeling pollution dispersion, population dynamics, and resource management.
   • Physics and Chemistry: For solving problems related to motion, energy, and rates of reactions.

   How to Start Learning Integrals

   Start by understanding the concept of the area under a curve and accumulate small sections to find the total. Begin with simple functions and gradually move to more complex scenarios. Practice makes perfect, and doing problems step-by-step will help solidify your understanding.

2. 2.Integration as an Inverse Process of Differentiation

   Integration as an Inverse Process of Differentiation

   Imagine you're watching a movie in reverse. Just as running the film backwards can help you trace the story from end to beginning, in mathematics, integration can be thought of as the reverse process of differentiation. Let's break this down with a simple example from everyday life.

   Everyday Analogy

   Suppose you're given a speedometer reading of a car, which shows how fast the car is going at every moment. Differentiation would involve figuring out how the speed changes moment by moment—like looking closely at the needle of the speedometer as it moves. Conversely, if you know how the speed changes and you want to find out the total distance traveled, you would integrate the speed over time. This means adding up all the little distances traveled in each tiny time interval.

   Mathematical Concept

   Differentiation gives you the rate at which a quantity changes, while integration sums up all these small changes to find out how much the quantity has changed over a period. In calculus, if 𝑓(𝑥)f(x) is a function, then its derivative 𝑓′(𝑥)f′(x) represents the rate of change of 𝑓f with respect to 𝑥x. Integrating 𝑓′(𝑥)f′(x), we retrieve the original function 𝑓(𝑥)f(x), perhaps plus a constant (since integration can only determine the function up to an additive constant).

   Real-life Applications

   1. Physics: Calculating the total displacement from velocity.
   2. Economics: Accumulating total revenue over time from a rate of revenue generation.
   3. Environmental Science: Integrating rate of pollutant emission to find total pollution.

   Careers and Industries

   Understanding the relationship between differentiation and integration is vital in fields such as:

   • Engineering: For designing systems where dynamics and changes over time are crucial.
   • Meteorology: For analyzing weather changes by integrating data over time.
   • Finance: For accumulating interest and analyzing investment growth.

   Steps to Understanding Integration as an Inverse of Differentiation

   1. Start with a function that represents a rate of change, such as velocity.
   2. Integrate this function over an interval to accumulate the total change, which might be distance traveled if the function was velocity.
   3. Understand the constant of integration: Since differentiation strips away constants (because the derivative of a constant is zero), integration adds an arbitrary constant called the "constant of integration".

   Activity to Try

   To see this in action, try differentiating a simple function like 𝑓(𝑥)=𝑥2f(x)=x2. You'll find that 𝑓′(𝑥)=2𝑥f′(x)=2x. Now, integrate 𝑓′(𝑥)f′(x) to see if you can get back to the original function 𝑓(𝑥)=𝑥2f(x)=x2, remembering that you may need to add a constant.

   Integration is not just a mathematical tool; it's a way of adding up all the small increments in which changes occur, giving us insight into how totals and accumulations are formed from rates of change. This makes it incredibly useful across many scientific and engineering disciplines.

3. 3.Properties of Indefinite Integrals

   Properties of Indefinite Integrals

   Indefinite integrals, also known as antiderivatives, have several properties that make them versatile tools in calculus. Let's explore some of these properties with proofs and examples.

   1. Linearity

   Property: The indefinite integral of a sum of functions is the sum of the indefinite integrals of the individual functions.

   Proof: Let 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) be two functions, and 𝐶C be a constant. Then, by the linearity of differentiation, we have:

   ∫[𝑓(𝑥)+𝑔(𝑥)] 𝑑𝑥=∫𝑓(𝑥) 𝑑𝑥+∫𝑔(𝑥) 𝑑𝑥∫[f(x)+g(x)]dx=∫f(x)dx+∫g(x)dx

   Example: ∫(3𝑥2+2𝑥+1) 𝑑𝑥=∫3𝑥2 𝑑𝑥+∫2𝑥 𝑑𝑥+∫1 𝑑𝑥∫(3x2+2x+1)dx=∫3x2dx+∫2xdx+∫1dx

   2. Constant Multiple

   Property: The indefinite integral of a constant times a function is equal to the constant times the indefinite integral of the function.

   Proof: Let 𝑓(𝑥)f(x) be a function, and 𝐶C be a constant. Then:

   ∫𝐶⋅𝑓(𝑥) 𝑑𝑥=𝐶⋅∫𝑓(𝑥) 𝑑𝑥∫C⋅f(x)dx=C⋅∫f(x)dx

   Example: ∫5𝑥2 𝑑𝑥=5⋅∫𝑥2 𝑑𝑥∫5x2dx=5⋅∫x2dx

   3. Power Rule

   Property: The indefinite integral of 𝑥𝑛xn with respect to 𝑥x is 𝑥𝑛+1𝑛+1+𝐶n+1xn+1​+C, where 𝑛≠−1n=−1 and 𝐶C is the constant of integration.

   Proof: Using the power rule of integration:

   ∫𝑥𝑛 𝑑𝑥=𝑥𝑛+1𝑛+1+𝐶∫xndx=n+1xn+1​+C

   Example: ∫𝑥3 𝑑𝑥=𝑥44+𝐶∫x3dx=4x4​+C

   4. Constant of Integration

   Property: When integrating, we always add a constant 𝐶C to the result to indicate that there could be many possible functions whose derivative is the original function.

   Example: ∫2𝑥 𝑑𝑥=𝑥2+𝐶∫2xdx=x2+C

   Example Problem:

   Problem: Find ∫(3𝑥2+2𝑥−5) 𝑑𝑥∫(3x2+2x−5)dx.

   Solution:

   Using the properties of indefinite integrals: ∫(3𝑥2+2𝑥−5) 𝑑𝑥=∫3𝑥2 𝑑𝑥+∫2𝑥 𝑑𝑥−∫5 𝑑𝑥∫(3x2+2x−5)dx=∫3x2dx+∫2xdx−∫5dx =3⋅𝑥33+2⋅𝑥22−5𝑥+𝐶=3⋅3x3​+2⋅2x2​−5x+C =𝑥3+𝑥2−5𝑥+𝐶=x3+x2−5x+C

   These properties make evaluating indefinite integrals more manageable, allowing us to find antiderivatives efficiently.

4. 4.Methods of Integration

   Integration is a fundamental concept in calculus, often used to find areas, volumes, solve differential equations, and in many real-world applications like physics and engineering. Let's explore some of the main methods of integration that you might find helpful.

   1. Substitution Method

   The substitution method is used when an integral contains a function and its derivative. You substitute a part of the integral to simplify it. This method is very similar to the reverse of the chain rule in differentiation.

   Example:
   Suppose you want to integrate ∫2𝑥cos⁡(𝑥2) 𝑑𝑥∫2xcos(x2)dx. Here, you can let 𝑢=𝑥2u=x2, then 𝑑𝑢=2𝑥 𝑑𝑥du=2xdx. This simplifies the integral to ∫cos⁡(𝑢) 𝑑𝑢∫cos(u)du, which is straightforward to integrate as sin⁡(𝑢)+𝐶sin(u)+C. Substituting back, you get sin⁡(𝑥2)+𝐶sin(x2)+C.

   2. Integration by Parts

   Integration by parts is based on the product rule of differentiation. It's useful when integrating the product of two functions.

   Formula:
   ∫𝑢 𝑑𝑣=𝑢𝑣−∫𝑣 𝑑𝑢∫udv=uv−∫vdu

   Example:
   To integrate ∫𝑥sin⁡(𝑥) 𝑑𝑥∫xsin(x)dx, choose 𝑢=𝑥u=x and 𝑑𝑣=sin⁡(𝑥) 𝑑𝑥dv=sin(x)dx. Then, 𝑑𝑢=𝑑𝑥du=dx and 𝑣=−cos⁡(𝑥)v=−cos(x). Applying the formula gives 𝑥(−cos⁡(𝑥))−∫−cos⁡(𝑥) 𝑑𝑥=−𝑥cos⁡(𝑥)+sin⁡(𝑥)+𝐶x(−cos(x))−∫−cos(x)dx=−xcos(x)+sin(x)+C.

   3. Partial Fractions

   This method is used when integrating rational functions (fractions where the numerator and denominator are polynomials). The idea is to break the fraction into simpler fractions that can be integrated individually.

   Example:
   To integrate ∫2𝑥2−1 𝑑𝑥∫x2−12​dx, you would decompose it into 𝐴𝑥−1+𝐵𝑥+1x−1A​+x+1B​. Solving for 𝐴A and 𝐵B and integrating the simpler fractions can make the task much easier.

   4. Trigonometric Substitution

   This method is helpful when dealing with integrals involving square roots of expressions like 𝑎2−𝑥2a2−x2, 𝑎2+𝑥2a2+x2, or 𝑥2−𝑎2x2−a2.

   Example:
   For ∫1−𝑥2 𝑑𝑥∫1−x2​dx, substituting 𝑥=sin⁡(𝜃)x=sin(θ) transforms the integral into a form involving sin⁡sin and cos⁡cos, which is easier to integrate.

   Real-World Applications

   These methods are not just academic; they have practical applications:

   • Engineering: Calculating the center of mass, electric and magnetic fields, or material properties.
   • Economics: Solving integral equations related to growth models or consumer surplus.
   • Physics: From motion equations in classical mechanics to wave functions in quantum mechanics.

   Integrating functions can seem daunting, but with practice, you'll find it an essential and satisfying part of solving mathematical problems. Try using these methods on different functions to see which ones simplify the task the best!

5. 5.Integration using Trigonometric Identities

   Integrating functions using trigonometric identities is a powerful technique, particularly when dealing with integrals that involve trigonometric functions. These identities can simplify complex expressions and make the integrals more manageable.

   Understanding Trigonometric Identities

   Trigonometric identities are equations involving trigonometric functions that are true for every value substituted into them. Some of the most commonly used identities in integration include:

   • Pythagorean Identity: sin⁡2(𝑥)+cos⁡2(𝑥)=1sin2(x)+cos2(x)=1
   • Double Angle Formulas: sin⁡(2𝑥)=2sin⁡(𝑥)cos⁡(𝑥)sin(2x)=2sin(x)cos(x), cos⁡(2𝑥)=cos⁡2(𝑥)−sin⁡2(𝑥)cos(2x)=cos2(x)−sin2(x)
   • Half Angle Formulas: sin⁡2(𝑥)=1−cos⁡(2𝑥)2sin2(x)=21−cos(2x)​, cos⁡2(𝑥)=1+cos⁡(2𝑥)2cos2(x)=21+cos(2x)​
   • Sum and Difference Formulas: For simplifying expressions like sin⁡(𝑎±𝑏)sin(a±b) or cos⁡(𝑎±𝑏)cos(a±b)

   These identities can be used to transform the integrand into a form that is easier to integrate.

   Example 1: Integrating Using the Pythagorean Identity

   Problem: Integrate ∫sin⁡2(𝑥) 𝑑𝑥∫sin2(x)dx.

   Solution:

   1. Use the half-angle identity: sin⁡2(𝑥)=1−cos⁡(2𝑥)2sin2(x)=21−cos(2x)​.
   2. Substitute into the integral:∫sin⁡2(𝑥) 𝑑𝑥=∫1−cos⁡(2𝑥)2 𝑑𝑥∫sin2(x)dx=∫21−cos(2x)​dx
   3. Distribute the integral:=12∫1 𝑑𝑥−12∫cos⁡(2𝑥) 𝑑𝑥=21​∫1dx−21​∫cos(2x)dx
   4. Integrate each term:=12𝑥−14sin⁡(2𝑥)+𝐶=21​x−41​sin(2x)+C

   Example 2: Using Double Angle Formulas

   Problem: Integrate ∫cos⁡2(𝑥) 𝑑𝑥∫cos2(x)dx.

   Solution:

   1. Use the half-angle identity: cos⁡2(𝑥)=1+cos⁡(2𝑥)2cos2(x)=21+cos(2x)​.
   2. Substitute into the integral:∫cos⁡2(𝑥) 𝑑𝑥=∫1+cos⁡(2𝑥)2 𝑑𝑥∫cos2(x)dx=∫21+cos(2x)​dx
   3. Distribute the integral:=12∫1 𝑑𝑥+12∫cos⁡(2𝑥) 𝑑𝑥=21​∫1dx+21​∫cos(2x)dx
   4. Integrate each term:=12𝑥+14sin⁡(2𝑥)+𝐶=21​x+41​sin(2x)+C

   Real-World Applications

   These techniques are not just theoretical; they have practical applications across various fields:

   • Physics: For computing work done by variable forces, or in oscillatory systems.
   • Engineering: In signal processing, especially in analyzing waves and vibrations.
   • Meteorology: In modeling wave patterns and atmospheric dynamics.

   Understanding and applying these identities in integration can significantly enhance your problem-solving skills in calculus, making complex integrals much easier to handle.

6. 6.Integrals of Some Particular Functions

   Integrating specific types of functions often involves recognizing patterns or applying specific techniques. Let’s explore how to integrate several common types of functions, providing examples for better understanding.

   1. Polynomials

   Integrating a polynomial involves applying the power rule of integration, which states: ∫𝑥𝑛 𝑑𝑥=𝑥𝑛+1𝑛+1+𝐶∫xndx=n+1xn+1​+C where 𝑛≠−1n=−1.

   Example: ∫𝑥3 𝑑𝑥=𝑥44+𝐶∫x3dx=4x4​+C

   2. Exponential Functions

   The integral of an exponential function is straightforward if it is of the form 𝑒𝑥ex: ∫𝑒𝑥 𝑑𝑥=𝑒𝑥+𝐶∫exdx=ex+C

   For a base other than 𝑒e, such as 𝑎𝑥ax, the integral is: ∫𝑎𝑥 𝑑𝑥=𝑎𝑥ln⁡(𝑎)+𝐶∫axdx=ln(a)ax​+C provided 𝑎>0a>0 and 𝑎≠1a=1.

   Example: ∫2𝑥 𝑑𝑥=2𝑥ln⁡(2)+𝐶∫2xdx=ln(2)2x​+C

   3. Trigonometric Functions

   Trigonometric functions have their own set of integration formulas:

   • Sine and Cosine: ∫sin⁡(𝑥) 𝑑𝑥=−cos⁡(𝑥)+𝐶∫sin(x)dx=−cos(x)+C ∫cos⁡(𝑥) 𝑑𝑥=sin⁡(𝑥)+𝐶∫cos(x)dx=sin(x)+C

   • Secant and Tangent: ∫sec⁡2(𝑥) 𝑑𝑥=tan⁡(𝑥)+𝐶∫sec2(x)dx=tan(x)+C ∫sec⁡(𝑥)tan⁡(𝑥) 𝑑𝑥=sec⁡(𝑥)+𝐶∫sec(x)tan(x)dx=sec(x)+C

   Example: ∫cos⁡(𝑥) 𝑑𝑥=sin⁡(𝑥)+𝐶∫cos(x)dx=sin(x)+C

   4. Logarithmic Functions

   The integral of the natural logarithm ln⁡(𝑥)ln(x) is: ∫ln⁡(𝑥) 𝑑𝑥=𝑥ln⁡(𝑥)−𝑥+𝐶∫ln(x)dx=xln(x)−x+C

   Example: ∫ln⁡(𝑥) 𝑑𝑥=𝑥ln⁡(𝑥)−𝑥+𝐶∫ln(x)dx=xln(x)−x+C

   5. Rational Functions

   Rational functions, or fractions involving polynomials in the numerator and denominator, can often be integrated by using partial fractions or other algebraic manipulations.

   Example: ∫1𝑥2−1 𝑑𝑥=12ln⁡∣𝑥−1𝑥+1∣+𝐶∫x2−11​dx=21​ln∣∣​x+1x−1​∣∣​+C

   Real-World Applications

   These integration techniques apply in various fields:

   • Physics: Calculating areas under curves, such as finding the work done by a force.
   • Engineering: Evaluating integrals for designing systems and components, such as in control engineering or electrical circuits.
   • Economics: Calculating consumer surplus and other economic measures that require integration.

   Understanding how to integrate these functions is crucial for solving many practical problems in science, engineering, and economics. With practice, you can develop a strong intuition for selecting the right technique for the right function.

7. 7.Integration by Partial Fractions

   Integration by partial fractions is a technique used to simplify and integrate rational functions, where the numerator and the denominator are polynomials. This method is particularly useful when the degree of the numerator is less than the degree of the denominator.

   Steps for Integration by Partial Fractions:

   1. Factorize the Denominator: Begin by factorizing the denominator into simpler, irreducible factors.
   2. Express as Partial Fractions: Write the rational function as a sum of simpler fractions, with constants in the numerators and the factors of the denominator in the denominators.
   3. Determine the Constants: Equate the original rational function to the sum of the partial fractions and solve for the constants in the numerators.
   4. Integrate Each Fraction: Finally, integrate each of the simpler fractions.

   Example 1:

   Integrate ∫3𝑥+2𝑥2−5𝑥+6 𝑑𝑥∫x2−5x+63x+2​dx.

   Step 1: Factorize the Denominator 𝑥2−5𝑥+6=(𝑥−2)(𝑥−3)x2−5x+6=(x−2)(x−3)

   Step 2: Express as Partial Fractions 3𝑥+2(𝑥−2)(𝑥−3)=𝐴𝑥−2+𝐵𝑥−3(x−2)(x−3)3x+2​=x−2A​+x−3B​

   Step 3: Determine the Constants Set up the equation: 𝐴(𝑥−3)+𝐵(𝑥−2)=3𝑥+2A(x−3)+B(x−2)=3x+2 Choosing convenient values for 𝑥x (like 𝑥=2x=2 and 𝑥=3x=3) makes it easy to solve: At 𝑥=2x=2: 𝐴(−1)=3(2)+2⇒𝐴=−8A(−1)=3(2)+2⇒A=−8 At 𝑥=3x=3: 𝐵(1)=3(3)+2⇒𝐵=11B(1)=3(3)+2⇒B=11

   Step 4: Integrate Each Fraction ∫−8𝑥−2 𝑑𝑥+∫11𝑥−3 𝑑𝑥=−8ln⁡∣𝑥−2∣+11ln⁡∣𝑥−3∣+𝐶∫x−2−8​dx+∫x−311​dx=−8ln∣x−2∣+11ln∣x−3∣+C

   Example 2:

   Integrate ∫𝑥𝑥2−4 𝑑𝑥∫x2−4x​dx.

   Step 1: Factorize the Denominator 𝑥2−4=(𝑥−2)(𝑥+2)x2−4=(x−2)(x+2)

   Step 2: Express as Partial Fractions 𝑥(𝑥−2)(𝑥+2)=𝐴𝑥−2+𝐵𝑥+2(x−2)(x+2)x​=x−2A​+x+2B​

   Step 3: Determine the Constants 𝐴(𝑥+2)+𝐵(𝑥−2)=𝑥A(x+2)+B(x−2)=x At 𝑥=2x=2: 𝐴(4)=2⇒𝐴=12A(4)=2⇒A=21​ At 𝑥=−2x=−2: 𝐵(−4)=−2⇒𝐵=12B(−4)=−2⇒B=21​

   Step 4: Integrate Each Fraction ∫12𝑥−2 𝑑𝑥+∫12𝑥+2 𝑑𝑥=12ln⁡∣𝑥−2∣+12ln⁡∣𝑥+2∣+𝐶∫x−221​​dx+∫x+221​​dx=21​ln∣x−2∣+21​ln∣x+2∣+C

   Example 3:

   Integrate ∫2𝑥+5𝑥2+𝑥−6 𝑑𝑥∫x2+x−62x+5​dx.

   Step 1: Factorize the Denominator 𝑥2+𝑥−6=(𝑥−2)(𝑥+3)x2+x−6=(x−2)(x+3)

   Step 2: Express as Partial Fractions 2𝑥+5(𝑥−2)(𝑥+3)=𝐴𝑥−2+𝐵𝑥+3(x−2)(x+3)2x+5​=x−2A​+x+3B​

   Step 3: Determine the Constants 𝐴(𝑥+3)+𝐵(𝑥−2)=2𝑥+5A(x+3)+B(x−2)=2x+5 At 𝑥=2x=2: 𝐴(5)=9⇒𝐴=95A(5)=9⇒A=59​ At 𝑥=−3x=−3: 𝐵(−5)=−1⇒𝐵=15B(−5)=−1⇒B=51​

   Step 4: Integrate Each Fraction ∫95𝑥−2 𝑑𝑥+∫15𝑥+3 𝑑𝑥=95ln⁡∣𝑥−2∣+15ln⁡∣𝑥+3∣+𝐶∫x−259​​dx+∫x+351​​dx=59​ln∣x−2∣+51​ln∣x+3∣+C

   This method systematically breaks down complex rational expressions into simpler parts that can be easily integrated, making it a valuable tool in calculus for solving integral problems involving rational functions.

8. 8.Integration by Parts

   Integration by parts is a technique derived from the product rule for differentiation and is useful for integrating products of functions. It can be particularly helpful when you deal with products of polynomials, exponentials, logarithms, or trigonometric functions.
   

   Formula for Integration by Parts:

   The formula is derived from the product rule for differentiation and states: ∫𝑢 𝑑𝑣=𝑢𝑣−∫𝑣 𝑑𝑢∫udv=uv−∫vdu Where:

   • 𝑢u and 𝑑𝑣dv are chosen from the integrand such that 𝑑𝑣dv is easily integrable.
   • After choosing 𝑢u and 𝑑𝑣dv, compute 𝑑𝑢du and 𝑣v (where 𝑣v is the integral of 𝑑𝑣dv).

   Example 1:

   Integrate ∫𝑥cos⁡(𝑥) 𝑑𝑥∫xcos(x)dx.

   Step 1: Choose 𝑢u and 𝑑𝑣dv Let 𝑢=𝑥u=x and 𝑑𝑣=cos⁡(𝑥) 𝑑𝑥dv=cos(x)dx. Then, compute 𝑑𝑢=𝑑𝑥du=dx and 𝑣=∫cos⁡(𝑥) 𝑑𝑥=sin⁡(𝑥)v=∫cos(x)dx=sin(x).

   Step 2: Apply the formula ∫𝑥cos⁡(𝑥) 𝑑𝑥=𝑥sin⁡(𝑥)−∫sin⁡(𝑥) 𝑑𝑥∫xcos(x)dx=xsin(x)−∫sin(x)dx =𝑥sin⁡(𝑥)+cos⁡(𝑥)+𝐶=xsin(x)+cos(x)+C

   Example 2:

   Integrate ∫𝑥2𝑒𝑥 𝑑𝑥∫x2exdx.

   Step 1: Choose 𝑢u and 𝑑𝑣dv Let 𝑢=𝑥2u=x2 and 𝑑𝑣=𝑒𝑥 𝑑𝑥dv=exdx. Then, compute 𝑑𝑢=2𝑥 𝑑𝑥du=2xdx and 𝑣=∫𝑒𝑥 𝑑𝑥=𝑒𝑥v=∫exdx=ex.

   Step 2: Apply the formula ∫𝑥2𝑒𝑥 𝑑𝑥=𝑥2𝑒𝑥−∫2𝑥𝑒𝑥 𝑑𝑥∫x2exdx=x2ex−∫2xexdx Continue with integration by parts for ∫2𝑥𝑒𝑥 𝑑𝑥∫2xexdx: Let 𝑢=2𝑥u=2x and 𝑑𝑣=𝑒𝑥 𝑑𝑥dv=exdx, then 𝑑𝑢=2 𝑑𝑥du=2dx and 𝑣=𝑒𝑥v=ex. ∫2𝑥𝑒𝑥 𝑑𝑥=2𝑥𝑒𝑥−∫2𝑒𝑥 𝑑𝑥=2𝑥𝑒𝑥−2𝑒𝑥∫2xexdx=2xex−∫2exdx=2xex−2ex Plugging back in: 𝑥2𝑒𝑥−(2𝑥𝑒𝑥−2𝑒𝑥)=𝑥2𝑒𝑥−2𝑥𝑒𝑥+2𝑒𝑥+𝐶x2ex−(2xex−2ex)=x2ex−2xex+2ex+C

   Real-World Applications

   Integration by parts is widely used in various fields such as:

   • Physics: To calculate work done when force depends on displacement.
   • Engineering: In signal processing and control systems.
   • Economics: In solving differential equations related to growth models.

9. 9.Some Properties of Definite Integrals

   Definite integrals have several key properties that make them particularly useful and interesting in mathematical analysis. Understanding these properties can help simplify the calculation of more complex integrals and solve problems more efficiently.

   Properties of Definite Integrals:

   1. Linearity

   The integral operator is linear, which means: ∫𝑎𝑏[𝑐𝑓(𝑥)+𝑑𝑔(𝑥)] 𝑑𝑥=𝑐∫𝑎𝑏𝑓(𝑥) 𝑑𝑥+𝑑∫𝑎𝑏𝑔(𝑥) 𝑑𝑥∫ab​[cf(x)+dg(x)]dx=c∫ab​f(x)dx+d∫ab​g(x)dx where 𝑐c and 𝑑d are constants. This property allows you to split the integral of a sum and factor out constants.

   2. Interval Additivity

   If 𝑐c is a point in the interval [𝑎,𝑏][a,b], then: ∫𝑎𝑏𝑓(𝑥) 𝑑𝑥=∫𝑎𝑐𝑓(𝑥) 𝑑𝑥+∫𝑐𝑏𝑓(𝑥) 𝑑𝑥∫ab​f(x)dx=∫ac​f(x)dx+∫cb​f(x)dx This property is useful for breaking down the integral into smaller sections, especially when dealing with complex functions or limits.

   3. Symmetry

   For even functions (𝑓(−𝑥)=𝑓(𝑥)f(−x)=f(x)), where the limits are symmetric about the origin: ∫−𝑎𝑎𝑓(𝑥) 𝑑𝑥=2∫0𝑎𝑓(𝑥) 𝑑𝑥∫−aa​f(x)dx=2∫0a​f(x)dx For odd functions (𝑓(−𝑥)=−𝑓(𝑥)f(−x)=−f(x)), the integral over symmetric limits cancels out: ∫−𝑎𝑎𝑓(𝑥) 𝑑𝑥=0∫−aa​f(x)dx=0

   4. Reversal of Limits

   Reversing the limits of integration changes the sign of the integral: ∫𝑎𝑏𝑓(𝑥) 𝑑𝑥=−∫𝑏𝑎𝑓(𝑥) 𝑑𝑥∫ab​f(x)dx=−∫ba​f(x)dx This is useful for adjusting the orientation or the order of integration limits.

   5. Zero Width Interval

   If the upper and lower limits are the same, the integral evaluates to zero: ∫𝑎𝑎𝑓(𝑥) 𝑑𝑥=0∫aa​f(x)dx=0 This property reflects the fact that an integral represents an "accumulated quantity," and no accumulation occurs over a zero-width interval.

   Examples:

   1. Using Linearity: ∫01(3𝑥2+2𝑥) 𝑑𝑥=3∫01𝑥2 𝑑𝑥+2∫01𝑥 𝑑𝑥∫01​(3x2+2x)dx=3∫01​x2dx+2∫01​xdx

   2. Using Symmetry (for an even function 𝑥2x2): ∫−11𝑥2 𝑑𝑥=2∫01𝑥2 𝑑𝑥∫−11​x2dx=2∫01​x2dx

10. 10.The Fundamental Theorem of Calculus

    The Fundamental Theorem of Calculus (FTC) is one of the most important concepts in calculus, bridging the two major branches: differential and integral calculus. It establishes the connection between differentiation and integration, providing a way to evaluate definite integrals without needing to compute sums of infinite series or limits of sums.

    The Fundamental Theorem of Calculus Consists of Two Parts:

    Part 1: Connection between Differentiation and Integration

    This part states that if 𝐹(𝑥)F(x) is an antiderivative of 𝑓(𝑥)f(x) over an interval [𝑎,𝑏][a,b], then the function 𝐹F is continuous on [𝑎,𝑏][a,b], differentiable on the open interval (𝑎,𝑏)(a,b), and 𝐹′(𝑥)=𝑓(𝑥)F′(x)=f(x) for all 𝑥x in (𝑎,𝑏)(a,b). It can be expressed as: 𝑑𝑑𝑥(∫𝑎𝑥𝑓(𝑡) 𝑑𝑡)=𝑓(𝑥)dxd​(∫ax​f(t)dt)=f(x)

    This means that the derivative of the integral of 𝑓f from a constant 𝑎a to a variable 𝑥x equals the integrand evaluated at 𝑥x.

    Part 2: Evaluation of Definite Integrals

    Part 2 of the FTC provides a practical method for calculating definite integrals. It states: ∫𝑎𝑏𝑓(𝑥) 𝑑𝑥=𝐹(𝑏)−𝐹(𝑎)∫ab​f(x)dx=F(b)−F(a) where 𝐹F is any antiderivative of 𝑓f, meaning 𝐹′(𝑥)=𝑓(𝑥)F′(x)=f(x).

    This tells us that the definite integral of 𝑓(𝑥)f(x) from 𝑎a to 𝑏b is the difference between the values of an antiderivative 𝐹F at 𝑏b and 𝑎a.

    Example Using the Fundamental Theorem of Calculus:

    Problem: Evaluate ∫0𝜋sin⁡(𝑥) 𝑑𝑥∫0π​sin(x)dx.

    Solution:

    1. Find an antiderivative of sin⁡(𝑥)sin(x): An antiderivative of sin⁡(𝑥)sin(x) is −cos⁡(𝑥)−cos(x).
    2. Apply the FTC Part 2: ∫0𝜋sin⁡(𝑥) 𝑑𝑥=(−cos⁡(𝜋))−(−cos⁡(0))=2∫0π​sin(x)dx=(−cos(π))−(−cos(0))=2

    Real-World Applications:

    • Physics: Calculating the area under velocity-time graphs to find total displacement.
    • Engineering: Evaluating the work done by a variable force along a path.
    • Economics: Finding the total consumer surplus from the demand curve.

11. 11.Evaluate Definite Integrals by Substitution

    Evaluating definite integrals using substitution is a technique that can simplify the integration process by transforming the integral into an easier one to calculate. This method is particularly useful when the integrand is a composite function or when a change of variables can simplify the bounds of integration.
    

    Steps to Evaluate Definite Integrals by Substitution:

    1. Identify a Substitution: Choose a substitution that simplifies the integrand. Let 𝑢=𝑔(𝑥)u=g(x), where 𝑔(𝑥)g(x) is a function that appears as part of the integrand.

    2. Differentiate to Find 𝑑𝑢du: Differentiate 𝑢u with respect to 𝑥x to find 𝑑𝑢du. This involves computing 𝑑𝑢=𝑔′(𝑥) 𝑑𝑥du=g′(x)dx.

    3. Change the Limits of Integration: When substituting, it's also necessary to change the limits of integration from 𝑥x values to 𝑢u values.

       • If the original limits are 𝑥=𝑎x=a and 𝑥=𝑏x=b, then the new limits will be 𝑢=𝑔(𝑎)u=g(a) and 𝑢=𝑔(𝑏)u=g(b).

    4. Substitute and Integrate: Replace 𝑑𝑥dx with 𝑑𝑢𝑔′(𝑥)g′(x)du​ and the 𝑥x-dependent expression with 𝑢u, then integrate with respect to 𝑢u.

    5. Back Substitute: After integrating, substitute back to return to the original variable if necessary.

    Example 1: Evaluate the Integral ∫0𝜋/4tan⁡(𝑥) 𝑑𝑥∫0π/4​tan(x)dx Using Substitution

    Step 1: Identify a Substitution Let 𝑢=tan⁡(𝑥)u=tan(x). This choice simplifies the integrand to 𝑢u.

    Step 2: Differentiate to Find 𝑑𝑢du 𝑑𝑢=sec⁡2(𝑥) 𝑑𝑥du=sec2(x)dx Since sec⁡2(𝑥)=1+tan⁡2(𝑥)sec2(x)=1+tan2(x), and with our substitution 𝑢=tan⁡(𝑥)u=tan(x), it becomes 𝑑𝑢=(1+𝑢2) 𝑑𝑥du=(1+u2)dx, or 𝑑𝑥=𝑑𝑢1+𝑢2dx=1+u2du​.

    Step 3: Change the Limits of Integration The original limits are 𝑥=0x=0 to 𝑥=𝜋/4x=π/4. At 𝑥=0x=0, 𝑢=tan⁡(0)=0u=tan(0)=0. At 𝑥=𝜋/4x=π/4, 𝑢=tan⁡(𝜋/4)=1u=tan(π/4)=1.

    Step 4: Substitute and Integrate ∫01𝑢1+𝑢2 𝑑𝑢∫01​1+u2u​du This integral can now be integrated directly: ∫01𝑢1+𝑢2 𝑑𝑢=12ln⁡(1+𝑢2)∣01=12ln⁡(2)−12ln⁡(1)=12ln⁡(2)∫01​1+u2u​du=21​ln(1+u2)∣∣​01​=21​ln(2)−21​ln(1)=21​ln(2)

    Example 2: Evaluate ∫14𝑑𝑥𝑥∫14​x​dx​

    Step 1: Identify a Substitution Let 𝑢=𝑥u=x​, hence 𝑥=𝑢2x=u2.

    Step 2: Differentiate to Find 𝑑𝑢du 𝑑𝑥=2𝑢 𝑑𝑢dx=2udu

    Step 3: Change the Limits of Integration Original limits are 𝑥=1x=1 to 𝑥=4x=4. At 𝑥=1x=1, 𝑢=1=1u=1​=1. At 𝑥=4x=4, 𝑢=4=2u=4​=2.

    Step 4: Substitute and Integrate ∫122𝑢 𝑑𝑢𝑢=∫122 𝑑𝑢=2𝑢∣12=4−2=2∫12​u2udu​=∫12​2du=2u∣∣​12​=4−2=2

12. 12.Definite Integrals

    Definite integrals have several key properties that make them particularly useful and interesting in mathematical analysis. Understanding these properties can help simplify the calculation of more complex integrals and solve problems more efficiently.
    

    Properties of Definite Integrals:

    1. Linearity

    The integral operator is linear, which means: ∫𝑎𝑏[𝑐𝑓(𝑥)+𝑑𝑔(𝑥)] 𝑑𝑥=𝑐∫𝑎𝑏𝑓(𝑥) 𝑑𝑥+𝑑∫𝑎𝑏𝑔(𝑥) 𝑑𝑥∫ab​[cf(x)+dg(x)]dx=c∫ab​f(x)dx+d∫ab​g(x)dx
    where 𝑐c and 𝑑d are constants. This property allows you to split the integral of a sum and factor out constants.

    2. Interval Additivity

    If 𝑐c is a point in the interval [𝑎,𝑏][a,b], then: ∫𝑎𝑏𝑓(𝑥) 𝑑𝑥=∫𝑎𝑐𝑓(𝑥) 𝑑𝑥+∫𝑐𝑏𝑓(𝑥) 𝑑𝑥∫ab​f(x)dx=∫ac​f(x)dx+∫cb​f(x)dx
    This property is useful for breaking down the integral into smaller sections, especially when dealing with complex functions or limits.

    3. Symmetry

    For even functions (𝑓(−𝑥)=𝑓(𝑥)f(−x)=f(x)), where the limits are symmetric about the origin: ∫−𝑎𝑎𝑓(𝑥) 𝑑𝑥=2∫0𝑎𝑓(𝑥) 𝑑𝑥∫−aa​f(x)dx=2∫0a​f(x)dx
    For odd functions (𝑓(−𝑥)=−𝑓(𝑥)f(−x)=−f(x)), the integral over symmetric limits cancels out: ∫−𝑎𝑎𝑓(𝑥) 𝑑𝑥=0∫−aa​f(x)dx=0

    4. Reversal of Limits

    Reversing the limits of integration changes the sign of the integral: ∫𝑎𝑏𝑓(𝑥) 𝑑𝑥=−∫𝑏𝑎𝑓(𝑥) 𝑑𝑥∫ab​f(x)dx=−∫ba​f(x)dx This is useful for adjusting the orientation or the order of integration limits.

    5. Zero Width Interval

    If the upper and lower limits are the same, the integral evaluates to zero: ∫𝑎𝑎𝑓(𝑥) 𝑑𝑥=0∫aa​f(x)dx=0 This property reflects the fact that an integral represents an "accumulated quantity," and no accumulation occurs over a zero-width interval.

    Examples:

    1. Using Linearity: ∫01(3𝑥2+2𝑥) 𝑑𝑥=3∫01𝑥2 𝑑𝑥+2∫01𝑥 𝑑𝑥∫01​(3x2+2x)dx=3∫01​x2dx+2∫01​xdx

    2. Using Symmetry (for an even function 𝑥2x2): ∫−11𝑥2 𝑑𝑥=2∫01𝑥2 𝑑𝑥∫−11​x2dx=2∫01​x2dx