Connect with us

calculus

Find the maximum and minimum values of a function

Published

on

Find the maximum and minimum values of the function. The maximum and minimum values of a function can be found by analyzing its behavior.

Steps to find maximum and minimum values

1. Determine the domain of a function:

  • Find where the function is defined. Check for points where the function might not exist (e.g., division by zero or square roots of negative numbers).

2. Find the critical points: 

  • Take the derivative of a function f(x) and set it equal to zero.

maximum and minimum of a function

Solve for x . These are the critical points.

  • Also find where f(x) is undefined within the domain, as these points can also indicate extrema.

3. Evaluate the endpoints (if applicable):

  • If the function is specified on a closed interval  [a,b], evaluate f(x) at the endpoints  x=a and x=b.

4. Perform the second derivative test (optional):

Compute f′′(x) at the critical points:

  • f ′′(x)>0 indicates that the point is a local minimum .
  • f′′(x)<0 indicates that the point is a local maximum.
  • If f′′(x)=0, the test is inconclusive, and other methods (like the first derivative test) are needed.

5. Compare values:

  • Plug the critical points and end points (if applicable) into the original function f(x) to find their corresponding values .
  • The minimum is the smallest value, and the maximum is the greatest.

How to find the maximum and minimum values of a function in a closed interval?

Example:

Find the maximum and minimum values of on [0, 3].

Step 1 : Domain

  • The polynomial function f(x) is defined for all real numbers. There are no concerns with the interval [0, 3].

Step 2: Critical Points

  • Consider the derivative:  f ′(x) = 3x^2 −6x.
  • Set f′(x)=0

3x^26x=03x(x2)=0x=0 or x=2.

critical points are x=0, x=2

Step 3: Endpoints

  • Evaluate f(x) at x=0 and x=3

Step 4: Compare values

  • ,

maximum and minimum of a function

Step 5: Conclusion

  • The maximum value is 4, appears at  x = 0 and  x = 3.
  • The minimum value is 0, which appears at x = 2.

 

 

 

Continue Reading

calculus

Definite integral properties

Published

on

By

Definite integral properties: Linearity , Reversal of limits, Zero width, Additivity, Bounds Inclusion, Comparison property, Absolute Value, Even and odd functions, Integration of a constant, Average value of a function, Additivity with zero contribution.

What is a definite integral?

A definite integral is a mathematical process that estimates the accumulation of quantities over an interval of time. The mathematical expression is as follows:

Here, and are the integration limits, is the function under consideration and represents an infinitesimally small change in. This process produces a number, which is commonly used to represent an area, volume, or other accumulated value.

Mathematical Representation

The definite integral of a function f(x) over the interval [a,b] is written as:

Here

  • a: The lower limit of integration.
  • b: The upper limit of integration.
  • f(x): The function to be integrated.
  • dx represents an infinitesimal change in .

The Geometric interpretation

One of the simplest and most straightforward ways to understand definite integrals is through geometric interpretation. Consider a curve defined by on a graph. The definite integral from to calculates the net area between the curve and the -axis over the interval.

  • Area above the x-axis: Contributions from add to the total.
  • Area above the y-axis: contributions from subtracting from the total.

Properties of definite integrals

1.  Linearity: Definite integral respects the principles of addition and scalar multiplication.

This means you can split the integral across sums or factor out constants.

2. Reversal of limits : Reversing the order of the limits of integration modifies the sign of the integral.

3. Zero width: The outcome is 0 if the integral’s upper and lower limits are equal.

This is because no area is accumulated over a zero-width interval.

4. Additivity (splitting intervals): A definite integral over an interval can be split into subintervals.

This is useful for breaking complex integrals into manageable parts.

5. Bounds Inclusion: If f(x) is continuous on [a,b], then:

Definite integral properties

This property is often used to handle piecewise functions.

6. Comparison property: If f(x) ≤ g(x) for any x in  [a,b], then:

Definite integral properties

If the functions are strictly ordered, the inequality becomes strict.

7. Absolute Value: The absolute value of an integral does not equal the integral of the absolute value in general.

Definite integral properties

This inequality reflects that the definite integral accounts for signed areas.

8. Even and odd functions

  • For an even function F(x)=F(-x) , integrated over symmetric limits

  • For an odd function F(x)=-F(-x), integrated over symmetric limits

Definite integral properties

9. Integration of a constant: If f(x) = c , where c is a constant, then:

This is the area of a rectangle whose width is b−a and its height is c.

10. Average value of a function: The definite integral can be used to compute the average value of f(x) over [a,b]

Definite integral properties

11. Additivity with zero contribution: If over an interval, then:

These are Definite integral properties.

Example 1:

Evaluate

Using linearity:

Definite integral properties

Compute each term

S0

Definite integral properties

Example 2:

Evaluate

Definite integral properties

Step 1: Analyze the problem

The integrand suggests that a substitution related to the square root Definite integral properties will simplify the integral . Let’s proceed with a trigonometric substitution.

Step 2: Sustitution

Let,

x=Sec(θ) , dx=Sec(θ) tan(θ) dθ 

The integral becomes

Definite integral properties

Simplify

Definite integral properties    Definite integral properties

Thus the integrand simplifies to

Definite integral properties

Step 3: Adjust limits of integration

When x=1 :

x= sec(θ) sec(θ)=1 θ=0

when x=4

x= sec(θ) sec(θ)=4 θ=sec−1(4)

The new limits are

Definite integral properties

Step 4: Integrate

The integral is now

The antiderivative of sec⁡^2(θ) is 

Definite integral properties

Step 5: Evaluate

Substitute the limits

Definite integral properties

and  at θ=0 tan(0)=0

Thus

 

Final Answer

Definite integral properties

Learn more indefinite integral

Continue Reading

calculus

Solve Indefinite integral- Math problems

Published

on

By

Solve Indefinite integral

Solve Indefinite integral :The indefinite integral of a function  f(x) is a family of functions  F(x) such that the derivative of  F(x) equals  f(x). In other words:

Solve Indefinite integral

The indefinite integral is written as:

Where

  • F(x) is the integrand ( the function being integrated)
  • dx indicates the variable of integration.
  • F(x) is the antiderivative of f(x).
  • The family of all potential antiderivatives is represented by the constant of integration, C.

Solve Indefinite integral math problems

Example 1

Solve Indefinite integral

Step 1: Expand sin3x using the angle formula

We are aware of the expansion for sin(3x) ⁡.

Substitute this in the integral

 

Solve Indefinite integralSolve Indefinite integral

Expand the integral

Solve Indefinite integral

Step 2: Solve the first term

For , use substitution: Let u=cos(x), so du=−sin⁡(x)dx This becomes:

Solve Indefinite integral

Substitute back u=cosx

Step 3: Solve the second term

For −4∫cos^⁡4(x)sin⁡3(x) dx, use

Solve Indefinite integral

Solve Indefinite integral

So the integral becomes

Substitute u=cos⁡(x), du=−sin⁡(x)dx:

Solve Indefinite integral

Simplify

Solve Indefinite integral

Integrate

Solve Indefinite integral

Substitute back u=cos(x)

Solve Indefinite integral

Step 4: Combine results

Add the two terms together

Solve Indefinite integral

Simplify

Final answer

Example 2

Solve Indefinite integral

Step 1: Use the half-angle identity

We know the identity

Substitute this into the integral

Simplify the square root

Solve Indefinite integral

Factor out √2

Step 2: Substitution

Let u=x/2, so x=2u and dx=2 du. Substitute:

Solve Indefinite integral

Simplify

Step 3: Integrate

The integral of cos (u) is sin(u)

Step 4: Back substitute

Recall that u=x/2, so:

Final answer

 

Example 3

Step 1: Use the product to sum identities

The product of two cosines can be expressed as

Apply this identity to cos2xcos3x:

Since cos⁡(−𝑥) = cos⁡𝑥 this can be simplified to:

Step 2: Substitute into the original integral

Now substitute this result into :

Simplify

Step 3: Solve term separately.

a) First term: ∫cos⁡xcos⁡5x dx

Use the product-to-sum identity again:

This becomes: as cos(−4x)=cos4x

Substitute into the integral:

Solve Indefinite integral

Evaluate

Solve Indefinite integral

Thus

Solve Indefinite integral

b) Second term: ∫cos²x dx

Use the identity cos²x= (1+cos2x)/2

Solve Indefinite integral

Evaluate

Solve Indefinite integral

Thus

Solve Indefinite integral

Step 4: Combine the results.

Combine the outcomes for the two terms now:

Solve Indefinite integral

Simplify

Solve Indefinite integralSolve Indefinite integral

Final answer

With practice, you’ll discover that solving integrals is a natural and gratifying process! Here are some basic integral rules

Some practice problems

1.Solve Indefinite integral

2.

3.Solve Indefinite integral

4.

5.Solve Indefinite integral

 

 

 

 

 

 

 

 

Continue Reading

calculus

Partial fraction formula, Example

Published

on

By

Partial fraction formula

Partial fraction formula: Integration by Partial Fractions is a method for integrating rational functions that are fractions of polynomials. The goal is to reduce a complex rational function to simpler fractions that are easier to integrate.

Steps for Partial Fraction Integration:

1.  Check to see if the degree of the numerator is less than that of the denominator:

If the degree of the numerator is greater than or equal to the degree of the denominator, use long division to rewrite the integral so that the degree of the numerator is less than that of the denominator.
2. Factor the denominator, if possible:

Try to break the denominator down into irreducible elements (linear or quadratic).
3. Set up the partial fractions: Depending on the factorization of the denominator, break the rational function into simpler fractions:

  • For linear factors (x-a) , use the form

Partial fraction formula

  • Use the form for repeated linear components  (x−a)^n

Partial fraction formula

  • For quadratic factors that cannot be reduced (x^2+bx+c) use the form

  • Use the following expression for repeated irreducible quadratic factors: (x^2 + bx + c)^n

4. Solve for the constants: After setting up the partial fractions , multiply both sides of the equation by the common denominator and equate coefficients of like powers of x to solve for the constants.

5. Integrate the simpler fractions: After determining the constants, integrate each term separately.

Partial fraction formula Partial fraction formula

1. Linear factors in the denominator

If the denominator consists of different linear factors:

Partial fraction formula

Example:

Steps

  1. Decompose

Partial fraction formula

2. Solve for A and B.

3. Integrate each term.

Partial fraction formula

Solution: A=1, B=1

Partial fraction formula

 

 

2. Repeated Linear Factors

If the denominator includes repeated linear factors:

Example:

Steps

  1. Decompose

Partial fraction formula

2. Integrate each term.

Solution:

3. Quadratic Factors

If the denominator includes an irreducible quadratic factor:

Partial fraction formula

Example:

  1. For irreducible x^2+4, rewrite:

Partial fraction formula

2. Solve for coefficients if needed (A = 0, B= 1)

3. Integrate using

Solution:

Partial fraction formula

4. Repeated Quadratic Factors

For several irreducible quadratic factors:

Example:

Partial fraction formula

Steps

  1. Decompose

2. Solve for coefficients.

3. Integrate each term.

Partial fraction formula

Solution:

 

5. Mixed Linear and Quadratic Factors

When the denominator contains both linear and quadratic factors:

Example:

Partial fraction formula

Steps

  1. Decompose

Partial fraction formula

2. Solve for A, B, and C.

3. Integrate each term separately.

 

Example :

Partial fraction formula

Step 1 : First, configure the partial fractions.

As (x+1)^2 (x−2), the denominator is already factored as (x+1)^2 (x−2). The integrand will therefore be represented as a sum of partial fractions of the following form:

Step 2: Multiply both sides by (𝑥+1)^2 ( 𝑥 − 2 ) 

Partial fraction formula

Step 3: Expand the right side.

Next, expand each phrase on the right-hand side:

Thus we get

Simplifying

 

 

Step 4: Compare the coefficients.

We equate the coefficients of powers of 𝑥 on both sides of the equation.

  • Coefficient of x^2: A+C=2
  • Coefficient of : A+B+2C=0
  • Constant term: −2A−2B+C=−1

Step 5: solve the system of equations.

We will now solve the system of equations:

  1. A+C=2
  2. -A +B+2C=0
  3. -2A-2B+2C=-1

From equation 1

C=2-A

Substitute C=2−A into equations (2) and (3):

From equation 2

Partial fraction formula

 

From equation 3

Partial fraction formula

Now solve equations (4) and (5).

Based on equation (4):

𝐵 = 3 𝐴 − 4

Substitute B=3A−4 into equation (5):

Partial fraction formula

Now substitute A=11/9 inti C = 2-A

Partial fraction formula

Finally , substitute A = 11/9 into B = 3A-4 :

Step 6: Write a partial fraction decomposition.

We currently have:

Partial fraction formula

Step 7: Integrate the terms.

We may now integrate each phrase separately.

Partial fraction formula

Partial fraction formula

Thus, the solution is:

Partial fraction formula

This is the final answer.

Practice problems

Solve the integral using partial fraction

1.

2.

3.

Partial fraction formula

 

4.

5.

 

 

Continue Reading

Title

Trending