Definite Integral - Examples related to special function

• In this lecture, we will focus on solving various examples related to special functions using definite integrals.
• We will discuss how to find the definite integral of functions involving trigonometric, logarithmic, and exponential functions.
• The definite integral is denoted by ∫ and represents the area under the curve between two given points on the x-axis.

Example 1: Definite Integral of a Trigonometric Function

• Find the value of ∫(0 to π/4) sin(x) dx.
• Solution:
  • Integrating sin(x) with respect to x gives -cos(x).
  • Evaluating the definite integral between 0 and π/4, we get -cos(π/4) - (-cos(0)).
  • Simplifying further, we have -√2/2 - (-1) = -√2/2 + 1.

Example 2: Definite Integral of a Logarithmic Function

• Evaluate ∫(1 to e) ln(x) dx.
• Solution:
  • We use the property that the integral of ln(x) is x ln(x) - x.
  • Substituting the limits, we have (e ln(e) - e) - (1 ln(1) - 1).
  • Since ln(1) = 0, the second term simplifies to 0.
  • Therefore, the value becomes e - e = 0.

Example 3: Definite Integral of an Exponential Function

• Calculate ∫(0 to 2) 3^x dx.
• Solution:
  • Integrating 3^x with respect to x gives (3^x) / ln(3).
  • Plugging in the limits, we get (3^2) / ln(3) - (3^0) / ln(3).
  • Simplifying further, we have 9/ln(3) - 1/ln(3) = 8/ln(3).

Example 4: Definite Integral of a Combination of Functions

• Find ∫(0 to π/2) cos(x) ln(sin(x)) dx.
• Solution:
  • The integral of cos(x) ln(sin(x)) involves two functions.
  • After evaluating the definite integral, the result is not in a standard form.
  • So, we will use integration by parts to simplify the expression.
• Remainder of the slides will be attached in the next response.

Slide 11: Integration by Parts

• Integration by parts is a technique used to solve the definite integral of a product of two functions.
• It is based on the product rule of differentiation.
• The formula for integration by parts is: ∫ u dv = uv - ∫ v du.

Slide 12: Applying Integration by Parts

• To apply integration by parts, we choose one function as u and the other as dv.
• We differentiate u to obtain du, and integrate dv to find v.
• Then we substitute these values into the formula and simplify the expression.

Slide 13: Example of Integration by Parts - ∫ x ln(x) dx

• Let’s solve the definite integral of x ln(x) dx using integration by parts.
• Choosing u = ln(x) and dv = x dx, we have du = (1/x) dx and v = (x^2)/2.
• Applying the integration by parts formula, we get: ∫ x ln(x) dx = (x^2 ln(x))/2 - ∫ (x^2/2) (1/x) dx.
• Simplifying further, we have: ∫ x ln(x) dx = (x^2 ln(x))/2 - ∫ (x/2) dx.

Slide 14: Continuation of Example - ∫ x ln(x) dx

• Continuing from the previous slide, we have: ∫ x ln(x) dx = (x^2 ln(x))/2 - (x^2)/4 + C.
• Therefore, the definite integral of x ln(x) dx is given by: ∫(a to b) x ln(x) dx = [(b^2 ln(b))/2 - (b^2)/4] - [(a^2 ln(a))/2 - (a^2)/4].

Slide 15: Example of Integration by Parts - ∫ x e^x dx

• Let’s solve the definite integral of x e^x dx using integration by parts.
• Choosing u = x and dv = e^x dx, we have du = dx and v = e^x.
• Applying the integration by parts formula, we get: ∫ x e^x dx = x e^x - ∫ e^x dx.
• Simplifying further, we have: ∫ x e^x dx = x e^x - e^x + C.

Slide 16: Definite Integral using Integration by Parts

• To find the definite integral using integration by parts, we need to evaluate the limits of integration.
• Let’s consider the example ∫(0 to 1) x e^x dx.
• Substituting the limits into the expression obtained in the previous slide, we have: ∫(0 to 1) x e^x dx = (1 e^1 - e^1) - (0 e^0 - e^0).
• Simplifying further, we get: ∫(0 to 1) x e^x dx = e - (1/e) + 1.

Slide 17: Example of Integration by Parts - ∫ x sin(x) dx

• Let’s solve the definite integral of x sin(x) dx using integration by parts.
• Choosing u = x and dv = sin(x) dx, we have du = dx and v = -cos(x).
• Applying the integration by parts formula, we get: ∫ x sin(x) dx = -x cos(x) - ∫ -cos(x) dx.
• Simplifying further, we have: ∫ x sin(x) dx = -x cos(x) + ∫ cos(x) dx.

Slide 18: Continuation of Example - ∫ x sin(x) dx

• Continuing from the previous slide, we have: ∫ x sin(x) dx = -x cos(x) + sin(x) + C.
• Therefore, the definite integral of x sin(x) dx is given by: ∫(a to b) x sin(x) dx = [-(b cos(b) - sin(b))] - [-(a cos(a) - sin(a))].

Slide 19: Importance of Definite Integrals

• Definite integrals are important in various fields, including physics, engineering, and economics.
• They help in calculating the area under curves, finding displacement, and solving problems involving rates of change.
• Understanding and applying definite integrals can enhance problem-solving skills and analytical thinking.

Slide 20: Summary

• In this lecture, we discussed examples related to special functions using definite integrals.
• We evaluated the definite integrals of trigonometric, logarithmic, and exponential functions.
• We also learned the technique of integration by parts to solve complicated integrals.
• Definite integrals are powerful tools in various fields and play a crucial role in mathematics and its applications.

Slide 21: Example of Definite Integral - ∫(0 to π/3) 2x cos(x^2) dx

• Evaluate the definite integral ∫(0 to π/3) 2x cos(x^2) dx.
• Solution:
  • To solve this integral, we can use a substitution method.
  • Let u = x^2, then du = 2x dx.
  • Substituting these values, the integral becomes ∫(0 to π/3) cos(u) du.
  • Integrate cos(u) to get sin(u).
  • Evaluating the integral with the limits, we have sin(x^2) evaluated from 0 to π/3.
  • Applying the limits, the value of the integral is sin((π/3)^2) - sin(0) = sin(π/9).

Slide 22: Example of Definite Integral - ∫(1 to 2) (2x + 1) dx

• Evaluate the definite integral ∫(1 to 2) (2x + 1) dx.
• Solution:
  • We can integrate each term separately and apply the limits.
  • Integrating 2x gives x^2, and integrating 1 gives x.
  • Applying the limits, the integral becomes [(2(2)^2 + 1) - (2(1)^2 + 1)].
  • Simplifying further, we have (8 + 1) - (2 + 1) = 8.

Slide 23: Example of Definite Integral - ∫(0 to π/4) sec^2(x) dx

• Evaluate the definite integral ∫(0 to π/4) sec^2(x) dx.
• Solution:
  • The integral of sec^2(x) is tan(x).
  • Applying the limits, the integral becomes tan(π/4) - tan(0).
  • Simplifying further, we have 1 - 0 = 1.

Slide 24: Example of Definite Integral - ∫(0 to π) csc(x) cot(x) dx

• Evaluate the definite integral ∫(0 to π) csc(x) cot(x) dx.
• Solution:
  • We can rewrite csc(x) cot(x) as (1/sin(x)) * (cos(x)/sin(x)).
  • The integral of (cos(x)/sin(x))^2 is -cot(x).
  • Applying the limits, the integral becomes -cot(π) - (-cot(0)).
  • Since cot(π) = 0, the value of the integral is -0 + 0 = 0.

Slide 25: Example of Definite Integral - ∫(0 to log(2)) e^x dx

• Evaluate the definite integral ∫(0 to log(2)) e^x dx.
• Solution:
  • The integral of e^x is e^x.
  • Applying the limits, the integral becomes e^log(2) - e^0.
  • Simplifying further, we have 2 - 1 = 1.

Slide 26: Example of Definite Integral - ∫(0 to 1) √(1 - x^2) dx

• Evaluate the definite integral ∫(0 to 1) √(1 - x^2) dx.
• Solution:
  • The integral represents the area under a half-circle of radius 1.
  • The area of a half-circle with radius 1 is π/4.
  • Therefore, the value of the integral is π/4.

Slide 27: Example of Definite Integral - ∫(0 to π/2) sec(x) dx

• Evaluate the definite integral ∫(0 to π/2) sec(x) dx.
• Solution:
  • Using the identity sec(x) = 1/cos(x), we can rewrite the integral as ∫(0 to π/2) (1/cos(x)) dx.
  • The integral of sec(x) is ln|sec(x) + tan(x)|.
  • Applying the limits, the integral becomes ln|sec(π/2) + tan(π/2)| - ln|sec(0) + tan(0)|.
  • Simplifying further, the value of the integral is ln(∞) - ln(1) = ∞.

Slide 28: Example of Definite Integral - ∫(0 to π) sin^2(x) dx

• Evaluate the definite integral ∫(0 to π) sin^2(x) dx.
• Solution:
  • Using the double angle identity, sin^2(x) can be written as (1 - cos(2x))/2.
  • The integral of (1 - cos(2x))/2 is (x - sin(2x)/4) evaluated between 0 and π.
  • Applying the limits, the integral becomes (π - sin(2π)/4) - (0 - sin(0)/4).
  • Simplifying further, the value of the integral is π - 0 = π.

Slide 29: Example of Definite Integral - ∫(0 to π) x^2 sin(x) dx

• Evaluate the definite integral ∫(0 to π) x^2 sin(x) dx.
• Solution:
  • To solve this integral, we can use integration by parts.
  • We choose u = x^2 and dv = sin(x) dx.
  • Differentiating u and integrating dv, we get du = 2x dx and v = -cos(x).
  • Applying the integration by parts formula, the integral becomes [-x^2 cos(x)] evaluated from 0 to π - ∫(-2x cos(x)) dx.
  • Integrating -2x cos(x), we get -2x sin(x) - 2 cos(x).
  • Applying the limits and simplifying, the value of the integral is π^2.

Slide 30: Summary

• In this lecture, we solved several examples of definite integrals involving special functions.
• We used techniques such as substitution, integration by parts, and trigonometric identities to evaluate the integrals.
• Definite integrals play a crucial role in calculus and have various applications in different fields.
• It is important to understand and practice solving different types of definite integrals to develop problem-solving skills.