Integral and Double Integral Calculus Examples with Solutions

Contents

Double Integral Calculus Questions with Step by Step Explanation

There are two main mathematical operators that you will encounter in calculus: integral and derivative. Both of these operators allow us to find specific information about a function. The process integration is nothing but to finding the anti-derivative of a function hence the anti-derivatives are also called the integrals of the function.

In this article, we’ll provide a brief overview of integral calculus and double integral calculus. Then, we’ll discuss how to evaluate the problems of integral and double integral through examples in step by step explanation.

What Is Integral Calculus?

Integral calculus is a branch of mathematics that deals with integrals. An integral is simply an area, volume, or length that has been calculated using the rules of integration. These rules are used to solve problems in calculus.

An integral can be thought of as a mathematical equation that relates one quantity to another quantity over time. The first quantity (the independent variable) changes while the second quantity (the dependent variable) remains constant.

What Is Double Integral?

In calculus, there is a powerful operation called the double integral. This operation can be used to calculate a variety of different geometric quantities, such as areas, volumes, and moments.

A double integral is a name given to the operation of finding the area of a two-dimensional region that is divided into multiple sub-regions. The integrand is a function that defines how the sub-regions are put together.

The limits of integration indicate the bounds of the region over which the integral is being calculated.

The integrand can be any function that takes two arguments: one for each sub-region. Any function that takes three or more arguments could be used as an integrand in a double integral calculation.

Once we have our function defined, we need to figure out how to integrate it. This involves solving for all possible values of ‘ x ‘ & ‘ y’ within each sub-region defined by our integrand function.

We call these solutions limits, and they indicate the bounds on our region over which we’re trying to calculate something specific (in this case area).

To find out how much volume is inside a container with an unknown shape, you can use double integration.

Integral and double integral examples with solutions

You can take assistance from the following examples to learn how to calculate the integral and double integral.

Example 1: For integral

Integrate the given function if the integrating variable of the function is “w”.

f(w) = 5w²– 6w³ + 2w⁴ + 5cos(w) – 20w

Solution

Step 1: Firstly, take the given function and write it in integral form by placing integration notation.

f(w) = 5w²– 6w³ + 2w⁴ + 5cos(w) – 20w

ʃ f(w) dw = ʃ [5w²– 6w³ + 2w⁴ + 5cos(w) – 20w] dw

Step 2: Now write the integration notation separately with each term of the function with the help of sum and difference rules of integral calculus.

ʃ [5w²– 6w³ + 2w⁴ + 5cos(w) – 20w] dw = ʃ [5w²] dw – ʃ [6w³] dw + ʃ [2w⁴] dw + ʃ [5cos(w)] dw – ʃ [20w] dw

Step 3: Take out the constant coefficients outside the integration notation.

ʃ [5w²– 6w³ + 2w⁴ + 5cos(w) – 20w] dw = 5ʃ [w²] dw – 6ʃ [w³] dw + 2ʃ [w⁴] dw + 5ʃ [cos(w)] dw – 20ʃ [w] dw

Step 4: Now apply the power and trigonometry rules of integral calculus to the above expression to integrate it.

= 5 [w²⁺¹/ 2 + 1] – 6 [w³⁺¹/ 3 + 1] + 2 [w⁴⁺¹/ 4 + 1] + 5 [sin(w)] – 20 [w¹⁺¹/ 1 + 1] + C

= 5 [w³/ 3] – 6 [w⁴/ 4] + 2 [w⁵/ 5] + 5 [sin(w)] – 20 [w²/ 2] + C

= 5/3 [w³] – 6/4 [w⁴] + 2/5 [w⁵] + 5 [sin(w)] – 20/2 [w²] + C

= 5/3 [w³] – 3/2 [w⁴] + 2/5 [w⁵] + 5 [sin(w)] – 10 [w²] + C

= 1.67 [w³] – 1.5 [w⁴] + 0.4 [w⁵] + 5 [sin(w)] – 10 [w²] + C

= 1.67w³– 1.5w⁴ + 0.4w⁵ + 5sin(w) – 10w² + C

This is why integrals are often considered to be difficult to solve. An antiderivative calculator can be used to reduce the difficulty of calculating the problems of integral calculus.

Example 2: For double integral

Integrate the given function if the integrating variables of the function are “x & y”.

f(x, y) = 2x²y – 3x³ + 4y³ + 2xy² – 6xy

Solution

Step 1: Firstly, take the given function and write it in double integral form by placing integration notation.

f(x, y) = 2x²y – 3x³ + 4y³ + 2xy² – 6xy

ʃ ʃ f(x, y) dx dy = ʃ ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx dy

Step 2: First of all, integrate the function with respect to “x”.

ʃ ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx dy = ʃ {ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx} dy … (1)

Step 3: Now take the x integrating value form 1 and write the integration notation separately with each term of the function with the help of sum and difference rules of integral calculus and take constant coefficients outside the notation.

ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx = ʃ [2x²y] dx – ʃ [3x³] dx + ʃ [4y³] dx + ʃ [2xy²] dx – ʃ [6xy] dx

= 2yʃ [x²] dx – 3ʃ [x³] dx + 4y³ʃ [1] dx + 2y²ʃ [x] dx – 6yʃ [x] dx

Step 4: Now apply the power and trigonometry rules of integral calculus to the above expression to integrate it.

= 2y [x²⁺¹/ 2 + 1] – 3 [x³⁺¹/ 3 + 1] + 4y³ [x] + 2y² [x¹⁺¹/ 1 + 1] – 6y [x¹⁺¹/ 1 + 1] + C

= 2y [x³/ 3] – 3 [x⁴/ 4] + 4y³ [x] + 2y² [x²/ 2] – 6y [x²/ 2] + C

= 2x³y/3 – 3x⁴/ 4 + 4y³x + 2y²x²/ 2 – 6yx²/ 2 + C

= 2x³y/3 – 3x⁴/ 4 + 4xy³ + x²y² – 3x²y + C

Step 5: Now place this value in (1).

ʃ ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx dy = ʃ {ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx} dy

= ʃ [2x³y/3 – 3x⁴/ 4 + 4xy³ + x²y² – 3x²y + C] dy … (2)

Step 6: Now integrate the above function with respect to “y”.

ʃ [2x³y/3 – 3x⁴/ 4 + 4xy³ + x²y² – 3x²y + C] dy = ʃ [2x³y/3] dy – ʃ [3x⁴/ 4] dy + ʃ [4xy³] dy + ʃ [x²y²] dy – ʃ [3x²y] dy + ʃ [C] dy

= 2x³/3ʃ [y] dy – 3x⁴/ 4ʃ [1] dy + 4x ʃ [y³] dy + x²ʃ [y²] dy – 3x²ʃ [y] dy + ʃ [C] dy

= 2x³/3 [y¹⁺¹ / 1 + 1] – 3x⁴/ 4 [y] + 4x [y³⁺¹ / 3 + 1] + x² [y²⁺¹ / 2 + 1] – 3x² [y¹⁺¹ / 1 + 1] + [Cy] + C

= 2x³/3 [y² / 2] – 3x⁴/ 4 [y] + 4x [y⁴ / 4] + x² [y³ / 3] – 3x² [y² / 2] + [Cy] + C

= 2x³y²/6 – 3x⁴y / 4 + 4xy⁴/4 + x²y³/3 – 3x²y²/2] + Cy + C

Hence,

ʃ ʃ [2x²y – 3x³ + 4y³ + 2xy² – 6xy] dx dy = 2x³y²/6 – 3x⁴y / 4 + 4xy⁴/4 + x²y³/3 – 3x²y²/2] + Cy + C

Sum Up

Integral calculus is the study of integrals, which are mathematical objects that can be used to describe certain areas or volumes. A double integral is a type of integration that allows for the calculation of two variables at once.

Related Articles

Coordinate geometry introduction | Locating of coordinate points

Frustum of cone formulas with examples

Centroid of the triangle formula with examples

Cone formulas – Surface Area and Volume

Surface Area and Volume of a Prism Formulas

Cylinder Formulas – Volume and Surface Area

Sphere, Hemisphere, Hollow Sphere Formulas – Surface Area and Volume

Cube and cuboid

Circle formulas in math

Quadrilateral formula for area and perimeter

Triangle Area Formulas