Integral Calculus - Example on Integral Inequality

  • We will solve an example involving integral inequality.
  • The integral inequality is a mathematical statement that relates the integral of a function to a constant or another integral.
  • The solution of the inequality involves finding the function and integrating it.

Example Problem

  • Let’s consider the following integral inequality: ∫[0 to 1] (x^2 - 1) dx ≤ ∫[0 to 1] (x^3 + 1) dx
  • Our task is to find the solution for this inequality.
  • We will start by finding the antiderivatives of both functions.

Antiderivative of (x^2 - 1)

  • To find the antiderivative of the function (x^2 - 1), we will use the power rule of integration.
  • Applying the power rule: ∫(x^2 - 1) dx = (1/3) x^3 - x + C
  • ‘C’ represents the constant of integration.

Antiderivative of (x^3 + 1)

  • To find the antiderivative of the function (x^3 + 1), we will again apply the power rule of integration.
  • Applying the power rule: ∫(x^3 + 1) dx = (1/4) x^4 + x + C
  • ‘C’ represents the constant of integration.

Rewriting the Integral Inequality

  • Now, let’s rewrite the original integral inequality using the found antiderivatives. (1/3) ∫[0 to 1] x^3 dx - ∫[0 to 1] 1 dx ≤ (1/4) ∫[0 to 1] x^4 dx + ∫[0 to 1] 1 dx

Evaluating the Indefinite Integrals

  • We will evaluate the definite integrals using the Fundamental Theorem of Calculus.
  • Applying the Fundamental Theorem of Calculus: (1/3) [x^4/4] - [x]^1_0 ≤ (1/4) [x^5/5] + [x]^1_0
  • Evaluating the definite integrals, we get: (1/3) (1/4) - 1/3 ≤ (1/4) (1/5) + 1/4

Simplifying the Inequality

  • Simplifying the inequality, we have: 1/12 - 1/3 ≤ 1/20 + 1/4
  • Further simplifying: -1/6 ≤ 1/20 + 1/4
  • Combining fractions: -1/6 ≤ 13/20

Solving the Inequality

  • To solve the inequality, we will compare the left-hand side and the right-hand side. -1/6 ≤ 13/20
  • Since the left-hand side is less than or equal to the right-hand side, the inequality holds true.

Conclusion

  • Our solution of the integral inequality is: ∫[0 to 1] (x^2 - 1) dx ≤ ∫[0 to 1] (x^3 + 1) dx -1/6 ≤ 13/20
  • Therefore, the given integral inequality is true.

Summary

  • In this example, we solved an integral inequality using the antiderivatives and evaluated the definite integrals.
  • We compared the left-hand side and the right-hand side to determine the validity of the inequality.
  • Integral inequalities are an important concept in calculus and have various applications in mathematical analysis.

Properties of Integrals

  • Integrals have several properties that help in their evaluation and understanding.
  • Some important properties of integrals include:
  1. Linearity: If f(x) and g(x) are integrable functions and a and b are constants, then the integral of af(x) + bg(x) is equal to a times the integral of f(x) plus b times the integral of g(x).
  1. Additivity: The integral of the sum of two functions is equal to the sum of their individual integrals.
  1. Change of Variable: The integral of a function can be simplified by performing a change of variable, which substitutes the variable of integration with a new variable.
  1. Substitution Rule: The integral of a composite function can be evaluated by using the substitution rule, which involves replacing the function inside the integral with a new function.
  1. Integration by Parts: The integral of a product of two functions can be evaluated by using the integration by parts method, which involves rewriting the integral as a combination of two new integrals.

Example: Integration by Parts

  • Let’s consider the integral ∫ x cos(x) dx.
  • To solve this integral, we can apply the integration by parts method.
  • Integration by parts formula: ∫ u dv = uv - ∫ v du
  • Let’s choose u = x and dv = cos(x) dx.

Example: Integration by Parts (Contd.)

  • Differentiating u, we get du = dx.
  • Integrating dv, we get v = sin(x).
  • Now, let’s apply the integration by parts formula: ∫ x cos(x) dx = x*sin(x) - ∫ sin(x) dx
  • The remaining integral can be easily evaluated as -cos(x).

Example: Integration by Parts (Contd.)

  • Plugging the values back into the original integral, we have: ∫ x cos(x) dx = x*sin(x) + cos(x) + C
  • Therefore, the solution to the integral ∫ x cos(x) dx is x*sin(x) + cos(x) + C, where C is the constant of integration.

Example: Change of Variable

  • Let’s consider the integral ∫ sin(2x) dx.
  • To simplify this integral, we can perform a change of variable.
  • Let’s choose u = 2x.
  • Differentiating u, we get du = 2 dx.
  • Rearranging the equation, we have du/2 = dx.

Example: Change of Variable (Contd.)

  • Now, let’s substitute the new variable into the integral: ∫ sin(2x) dx = ∫ sin(u) (du/2)
  • Simplifying, we have: (1/2) ∫ sin(u) du
  • The integral of sin(u) is -cos(u).

Example: Change of Variable (Contd.)

  • Plugging the values back into the integral, we have: (1/2) (-cos(u)) + C
  • Since u = 2x, the final solution becomes: (-1/2) cos(2x) + C
  • Therefore, the solution to the integral ∫ sin(2x) dx is (-1/2) cos(2x) + C, where C is the constant of integration.

Summary

  • Integrals have various properties that make them useful in solving mathematical problems.
  • Some important properties include linearity, additivity, change of variable, substitution rule, and integration by parts.
  • These properties help simplify integrals and make them easier to evaluate.
  • In the examples given, we used integration by parts and change of variable to solve integrals.

Summary (Contd.)

  • Integration by parts involves splitting the integral of a product into two new integrals.
  • Change of variable allows us to substitute the variable of integration with a new variable to simplify the integral.
  • It is essential to understand these properties and techniques to solve complex integrals effectively.
  • Practicing these methods and understanding their applications will help in scoring well in the exams.

Questions

  • Let’s practice some problems related to integrals and their properties.
  • Solve the following integrals using the techniques discussed in this lecture:
  1. ∫ e^x sin(x) dx
  1. ∫ (2x + 5) cos(x^2) dx
  1. ∫ x^3 e^x dx
  1. ∫ ln(x)/x dx
  • Take your time to solve these problems and ask any questions if you face difficulties.

Example: ∫ e^x sin(x) dx

  • To solve this integral, we can use integration by parts method.
  • Let’s choose u = e^x and dv = sin(x) dx.
  • Differentiating u, we get du = e^x dx.
  • Integrating dv, we get v = -cos(x).

Example: ∫ e^x sin(x) dx (Contd.)

  • Now, let’s apply the integration by parts formula: ∫ e^x sin(x) dx = -e^x cos(x) - ∫ -cos(x) e^x dx
  • Simplifying, we have: ∫ e^x sin(x) dx = -e^x cos(x) + ∫cos(x) e^x dx
  • It is convenient to solve this integral using the method of integration by parts again.

Example: ∫ e^x sin(x) dx (Contd.)

  • Let’s choose u = e^x and dv = cos(x) dx.
  • Differentiating u, we get du = e^x dx.
  • Integrating dv, we get v = sin(x).
  • Applying the integration by parts formula again: ∫ e^x sin(x) dx = -e^x cos(x) + ∫ cos(x) e^x dx = -e^x cos(x) + e^x sin(x) - ∫ sin(x) e^x dx

Example: ∫ e^x sin(x) dx (Contd.)

  • Rearranging, we have: 2∫ e^x sin(x) dx = -e^x cos(x) + e^x sin(x)
  • Dividing by 2 on both sides: ∫ e^x sin(x) dx = (-e^x cos(x) + e^x sin(x))/2 + C
  • Therefore, the solution to the integral ∫ e^x sin(x) dx is (-e^x cos(x) + e^x sin(x))/2 + C, where C is the constant of integration.

Example: ∫ (2x + 5) cos(x^2) dx

  • To solve this integral, we can use the substitution method.
  • Let’s choose u = x^2.
  • Differentiating u, we get du = 2x dx.

Example: ∫ (2x + 5) cos(x^2) dx (Contd.)

  • Rearranging the equation, we have: (1/2) du = x dx
  • Now, let’s substitute the new variable into the integral: ∫ (2x + 5) cos(x^2) dx = ∫ (2x + 5) cos(u) (1/2) du
  • Simplifying, we have: ∫ (x + 5/2) cos(u) du

Example: ∫ (2x + 5) cos(x^2) dx (Contd.)

  • Expanding the integral, we get: ∫ x cos(u) du + (5/2) ∫ cos(u) du
  • The first integral can be easily evaluated as sin(u).
  • The second integral is equal to (5/2) sin(u) + C.

Example: ∫ (2x + 5) cos(x^2) dx (Contd.)

  • Plugging the values back into the original integral, we have: ∫ (2x + 5) cos(x^2) dx = sin(u) + (5/2) sin(u) + C = (7/2) sin(u) + C
  • Since u = x^2, the final solution becomes: (7/2) sin(x^2) + C
  • Therefore, the solution to the integral ∫ (2x + 5) cos(x^2) dx is (7/2) sin(x^2) + C, where C is the constant of integration.

Example: ∫ x^3 e^x dx

  • To solve this integral, we can use the method of integration by parts.
  • Let’s choose u = x^3 and dv = e^x dx.
  • Differentiating u, we get du = 3x^2 dx.
  • Integrating dv, we get v = e^x.

Example: ∫ x^3 e^x dx (Contd.)

  • Applying the integration by parts formula: ∫ x^3 e^x dx = x^3 e^x - ∫ 3x^2 e^x dx
  • Integrating the remaining integral by parts: ∫ 3x^2 e^x dx = 3x^2 e^x - ∫ 6x e^x dx
  • Applying integration by parts again: ∫ 6x e^x dx = 6x e^x - ∫ 6e^x dx
  • The integral of 6e^x is simply 6e^x itself.