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Solve the area of conics using integration

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Integration is a useful method in calculus for finding areas under curves. Integration can help identify the exact area encompassed by conic section-described curves like circles, parabolas, and hyperbolas. In this blog, we’ll look at the general rules and techniques for calculating the areas under the curves of these fundamental shapes. Solve the area of conics using integration.

Solve the area of conics using integration

1. Find the area of a circle: One of the simplest and most well-known geometric shapes is a circle. In Cartesian coordinates, a circle’s standard equation is:

Solve the area of conics using integration

Where r is the radius of the circle.

We calculate y using the circle’s equation:

To calculate the area of the upper half of the circle, first integrate the positive component of y and then double it to account for both parts. The integral of the area of the upper half of the circle is:

Solve the area of conics using integration

This integral must be evaluated using a standard formula or a trigonometric substitution (such as x=r sin(θ)). The outcome will provide:

2. Find the area of a parabola: A quadratic equation defines another common curve, a parabola. In Cartesian coordinates, a parabola takes the following standard form:

Solve the area of conics using integration

To find the area under a parabola, normally integrate the function over the limits where it meets the x-axis.

Method 1: Area under one side of a parabola

For a simple upward opening parabola like y=x^2, we can calculate the area under the curve from x=-a to x=a as:

Solve the area of conics using integration

This yields

Method 2: Area between two parabolas

To find the area between two parabolas, integrate the difference between y 1​=ax^2 +bx + c and y 2​=dx^2 +ex +f.

Solve the area of conics using integration

where the intersections of the parabolas are denoted by x 1 and x 2. These points can be obtained by solving y 1 = y 2, which is to say, finding the roots of the equation ax^2 + bx + c = dx^2 + ex + f.

3. Find the area of a hyperbola:

A hyperbola is a curve defined by an equation of the form:

This represents a standard horizontal hyperbola.

Solving for y, we get

To calculate the area under the upper branch of a hyperbola, integrate from x=a to x=b.

Solve the area of conics using integration

This integral requires a trigonometric substitution, such as x =, and evaluates to the desired area.

4. Find the area of the ellipse:

The standard equation of ellipse is

Where a and b are the semimajor and semiminor axes, respectively.

we can solve y in terms of x:

To find the area under the upper half of the ellipse, integrate the positive root from x=−a to x=a:

Solve the area of conics using integration

This integral can be solved using a trigonometric substitution, such as x=a sin⁡(θ)x , and the result is:

The overall area of the ellipse is twice that of the upper half.

Calculus requires the ability to solve for the area bounded by conic sections using integration. Regardless of whether you’re working with a circle, parabola, hyperbola, or ellipse, the general method entails:

  • Setting up the equation of the curve.
  • Identifying the appropriate limits of integration.
  • Using integration to compute the area

Mastering these approaches enables you to calculate areas for various conic sections, which is useful in a variety of fields including physics, engineering, and mathematics. So try to Solve the area of conics using integration .

 

 

 

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Definite integral properties

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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

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Find the maximum and minimum values of a function

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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.

 

 

 

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Solve Indefinite integral- Math problems

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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

 

 

 

 

 

 

 

 

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