Indefinite Integral - INTEGRALS

• In mathematics, an indefinite integral is the most general antiderivative that is obtained from a given function.
• It is denoted as ∫f(x) dx, where f(x) is the function and dx represents the variable of integration.
• The indefinite integral is a fundamental concept in calculus and is closely related to the derivative.

Basic Rules for Indefinite Integration

1. Constant Rule: ∫c dx = cx + C, where c is a constant and C is the constant of integration.
   • Example: ∫3 dx = 3x + C

2. Power Rule: ∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1.
   • Example: ∫x^2 dx = (x^3)/3 + C

3. Sum and Difference Rule: ∫(f(x) ± g(x)) dx = ∫f(x)dx ± ∫g(x)dx
   • Example: ∫(3x^2 - 5x) dx = ∫3x^2 dx - ∫5x dx = x^3 - (5/2)x^2 + C

Basic Rules for Indefinite Integration (continued)

4. Constant Multiple Rule: ∫c * f(x) dx = c * ∫f(x) dx, where c is a constant.
   • Example: ∫5x^3 dx = 5 * ∫x^3 dx = (5/4) * x^4 + C

5. Exponential Rule: ∫e^x dx = e^x + C
   • Example: ∫e^2x dx = (1/2) * e^(2x) + C

6. Trigonometric Rule: ∫sin(x) dx = -cos(x) + C
   • Example: ∫sin(3x) dx = (-1/3) * cos(3x) + C

Methods of Integration

1. Direct Method: We can directly use the basic rules to find the indefinite integral of a given function.
   • Example: ∫(2x + 3) dx = x^2 + 3x + C

2. Integration by Substitution: This method involves substituting a variable to simplify the integral.
   • Example: ∫2x * (x^2 + 1)^3 dx. Let u = x^2 + 1, then du/dx = 2x. The integral becomes ∫u^3 du, which is easy to integrate.

Methods of Integration (continued)

3. Integration by Parts: This method involves applying the product rule in reverse.
   • Example: ∫x * sin(x) dx. Let u = x and dv/dx = sin(x). Integrating by parts, we have ∫u * dv = u * v - ∫v * du.

4. Integration by Partial Fractions: This method is used to integrate rational functions by decomposing them into simpler fractions.
   • Example: ∫(3x + 2)/(x^2 - x) dx. We can use partial fractions to rewrite the function as A/(x - 1) + B/x.

Common Indefinite Integrals

1. ∫(dx) = x + C

2. ∫(k dx) = kx + C

3. ∫(x^n dx) = (x^(n+1))/(n+1) + C, n ≠ -1

4. ∫(e^x dx) = e^x + C

5. ∫(a^x dx) = (1/ln(a)) * a^x + C, a > 0, a ≠ 1

Common Indefinite Integrals (continued)

6. ∫(sin(x) dx) = -cos(x) + C

7. ∫(cos(x) dx) = sin(x) + C

8. ∫(tan(x) dx) = -ln|cos(x)| + C

9. ∫(sec^2(x) dx) = tan(x) + C

10. ∫(cot(x) dx) = -ln|sin(x)| + C

Common Indefinite Integrals (continued)

11. ∫(csc^2(x) dx) = -cot(x) + C

12. ∫(1/(1 + x^2) dx) = tan^(-1)(x) + C

13. ∫(1/sqrt(1 - x^2) dx) = sin^(-1)(x) + C

14. ∫(1/x dx) = ln|x| + C

15. ∫(sec(x) * tan(x) dx) = sec(x) + C

Additional Techniques in Integration

1. Integration by Trigonometric Substitution: This method is useful when dealing with integrals involving square roots of quadratic expressions.
   • Example: ∫sqrt(x^2 + 1) dx. Substitute x = tanθ to simplify the integral.

2. Integration by Completing the Square: This method involves manipulating the integrand to convert it into a perfect square.
   • Example: ∫(x^2 + 4x + 3) dx. Complete the square to rewrite the integrand as (x + 2)^2 - 1.

Additional Techniques in Integration (continued)

3. Integration by Rationalizing Substitutions: This method is used to simplify integrals by applying trigonometric substitutions to expressions involving square roots.
   • Example: ∫sqrt(4 - x^2) dx. Substitute x = 2sinθ to simplify the integral.

4. Integration by Tables: This method involves referring to a pre-defined table of integrals to find the integral of a given function.

11. Integration by Trigonometric Substitution

• When dealing with integrals involving square roots of quadratic expressions, this method can be helpful.
• Example: ∫sqrt(x^2 + 1) dx. Substitute x = tanθ to simplify the integral.
• By substituting x = tanθ, dx = sec^2(θ) dθ.
• The integral becomes ∫sec(θ) dθ.
• Using the trigonometric identity sec^2(θ) = tan^2(θ) + 1, we simplify further.
• The integral now becomes ∫sqrt(tan^2(θ) + 1) sec(θ) dθ.
• Simplifying, we get ∫sqrt(sec^2(θ)) sec(θ) dθ.
• The integral becomes ∫sec(θ) sec(θ) dθ.
• By employing the identity sec^2(θ) = 1 + tan^2(θ), we simplify the integral even further.
• Finally, we have ∫sec^2(θ) dθ.

12. Integration by Completing the Square

• This method involves manipulating the integrand to convert it into a perfect square.
• Example: ∫(x^2 + 4x + 3) dx. Complete the square to rewrite the integrand as (x + 2)^2 - 1.
• The integrand x^2 + 4x + 3 can be rewritten as (x + 2)^2 - 1.
• The integral becomes ∫((x + 2)^2 - 1) dx.
• Using the distributive property, we split the integral into two parts.
• We get ∫(x + 2)^2 dx - ∫1 dx.
• Applying the power rule, we integrate (x + 2)^2 to get (x + 2)^3/3.
• Simplifying, the integral becomes (x + 2)^3/3 - x + C.

13. Integration by Rationalizing Substitutions

• This method is used to simplify integrals by applying trigonometric substitutions to expressions involving square roots.
• Example: ∫sqrt(4 - x^2) dx. Substitute x = 2sinθ to simplify the integral.
• By substituting x = 2sinθ, dx = 2cosθ dθ.
• The integral becomes ∫sqrt(4 - (2sinθ)^2) * 2cosθ dθ.
• Simplifying, we have ∫sqrt(4 - 4sin^2θ) * 2cosθ dθ.
• Applying the trigonometric identity sin^2θ + cos^2θ = 1, we can simplify further.
• The integral now becomes ∫sqrt(4cos^2θ) * 2cosθ dθ.
• Simplifying, we get ∫2cos^2θ * cosθ dθ.
• Using the identity cos^2θ = (1 + cos2θ)/2, we further simplify the integral.
• Finally, we have ∫cos^3θ dθ.

14. Integration by Tables

• This method involves referring to a pre-defined table of integrals to find the integral of a given function.
• Various mathematical handbooks and resources provide tables of integrals for different types of functions.
• To use the table, it is important to identify the form of the integral and its corresponding entry in the table.
• The table provides the integral in terms of functions or combinations of functions.
• Example: ∫x^3e^x dx can be evaluated using the table of integrals.
• To find the integral, look for the entry that matches the form of the integrand.
• In this case, the table may provide the integral as a combination of polynomial and exponential functions.

15. Integration by Parts

• Integration by parts involves applying the product rule in reverse to find an indefinite integral.
• The product rule states that the derivative of a product of two functions f(x) and g(x) is given by (f(x)g’(x) + f’(x)g(x)).
• The formula for integration by parts is: ∫f(x)g’(x) dx = f(x)g(x) - ∫f’(x)g(x) dx.
• Example: ∫x * sin(x) dx. Let u = x and dv/dx = sin(x).
• Differentiating u with respect to x, we get du/dx = 1.
• Integrating dv/dx with respect to x, we get v = -cos(x).
• Applying the integration by parts formula, we have ∫x * sin(x) dx = -x * cos(x) - ∫(-cos(x)) dx.
• Simplifying further, the integral becomes -x * cos(x) + sin(x) + C.
• The constant of integration, C, accounts for the difference between the indefinite integral and the original function.

16. Improper Integrals

• Improper integrals are integrals that either have infinite limits or where the function being integrated is not defined for some values within the limits.
• They involve evaluating integrals where the limits of integration are infinite or where the integrand has a singularity within the limits.
• There are two types of improper integrals: Type 1 - Infinite limits and Type 2 - Discontinuous functions.
• For Type 1 improper integrals, a limit is taken as one or both of the integration limits approach infinity.
• For Type 2 improper integrals, the integrand is not defined or is discontinuous at one or more points within the limits.
• Example: ∫(1/x) dx over the interval (1, ∞) is a Type 1 improper integral.
• In this case, the limit is taken as the lower limit approaches 1 and the upper limit approaches infinity.

17. Applications of Indefinite Integrals - Area Under a Curve

• Indefinite integrals can be used to find the area under a curve.
• The area under a curve between two points can be determined by evaluating the indefinite integral of the function over the interval between those points.
• Example: Find the area under the curve y = x^2 between x = 1 and x = 3.
• We need to evaluate ∫(x^2) dx over the interval [1, 3].
• By integrating x^2, we get (x^3)/3 + C.
• Evaluating the integral at the upper and lower limits, we have [(3^3)/3 - (1^3)/3].
• Simplifying further, the area under the curve is (27/3 - 1/3), which equals 8.

18. Applications of Indefinite Integrals - Volume of Solids of Revolution

• Indefinite integrals can also be used to find the volume of solids of revolution.
• A solid of revolution is obtained by rotating a curve about a line or axis.
• Example: Find the volume of the solid generated when the curve y = x^2 is revolved about the x-axis between x = 0 and x = 2.
• We need to evaluate the integral ∫(pi * (x^2)^2) dx over the interval [0, 2].
• Simplifying, the integral becomes ∫(pi * x^4) dx.
• By integrating x^4, we get (pi * (x^5)/5 + C.
• Evaluating the integral at the upper and lower limits, we have [(pi * (2^5)/5 - (pi * (0^5)/5].
• Simplifying further, the volume of the solid is (64pi/5 - 0), which equals (64pi/5).

19. Applications of Indefinite Integrals - Work and Fluid Force

• Indefinite integrals can be used to calculate work and fluid force.
• Work is the product of force and displacement, and the indefinite integral can be used to find the work done by a variable force.
• Fluid force refers to the force exerted by a fluid on a surface and can also be evaluated using an indefinite integral.
• Example: Calculate the work done by a force F(x) = 3x^2 on an object displaced from x = 1 to x = 5.
• We need to evaluate the integral ∫(3x^2) dx over the interval [1, 5].
• By integrating 3x^2, we get x^3 + C.
• Evaluating the integral at the upper and lower limits, we have [(5^3 + C) - (1^3 + C)].
• Simplifying further, the work done is (125 - 1), which equals 124.

20. Applications of Indefinite Integrals - Center of Mass

• Indefinite integrals are also used to find the center of mass of an object or system.
• The center of mass is the average position of all the parts of the object or system, weighted according to their masses.
• The integral can be used to find the center of mass by dividing the total mass into infinitesimally small parts and calculating their respective distances from the origin.
• Example: Find the center of mass of a thin wire of density λ(x) = 3x located along the interval [0, 2].
• We need to evaluate the integral ∫(xλ(x)) dx over the interval [0, 2].
• By integrating xλ(x), we get (x^2)/2 + C.
• Evaluating the integral at the upper and lower limits, we have [(2^2)/2 + C - (0^2)/2 + C].
• Simplifying further, the center of mass is (2 - 0), which equals 2. <!– Slide 21 –>

Integration by Substitution - Method

• Integration by substitution is a method used to simplify integrals by making a substitution of variables.
• It allows us to rewrite the integral in terms of a new variable, making it easier to evaluate.
• The general form of integration by substitution is: ∫f(g(x))g’(x)dx = ∫f(u)du, where u = g(x).
• Steps for Integration by Substitution:
  1. Identify a part of the integrand that fits the form u = g(x).
  2. Calculate du/dx, the derivative of u with respect to x.
  3. Rewrite the integral in terms of u and du, replacing the integrand and dx.
  4. Evaluate the new integral with respect to u.
  5. Substitute back the original variable x in the final answer. <!– Slide 22 –>

Integration by Substitution - Example

• Example: Evaluate the integral ∫(2x + 1)^5dx.
• Solution:
  1. Let u = 2x + 1.
  2. Calculate du/dx: du/dx = 2.
  3. Rewrite the integral in terms of u and du: ∫u^5 * (1/2)du.
  4. Evaluate the integral with respect to u: (1/2) * ∫u^5du = (1/2) * (u^6/6) + C.
  5. Substitute back the original variable x: (1/2) * ((2x + 1)^6/6) + C.
  • Final Answer: (1/12) * (2x + 1)^6 + C. <!– Slide 23 –>

Integration by Parts - Method

• Integration by parts is a method used to integrate the product of two functions.
• The general form of integration by parts is: ∫u * dv = u * v - ∫v * du.
• Steps for Integration by Parts:
  1. Choose u and dv from the given function.
  2. Calculate du and v using the derivatives and integrals of u and dv.
  3. Apply the integration by parts formula and evaluate the integral on the right side.
  4. Simplify the resulting integral if possible.
  5. Repeat the process if necessary, until the integral is easily computable. <!– Slide 24 –>

Integration by Parts - Example

• Example: Evaluate the integral ∫x * ln(x) dx.
• Solution:
  1. Choose u = ln(x) and dv = x dx.
  2. Calculate du and v:
     • du = (1/x) dx,
     • v =