In mathematics, the calculus is a widely used branch that deals with the properties & behavior of functions. It has a wide range of applications and uses in economics, physics, engineering, & computer science.

Solving calculus problems is an integral component of mathematics that can be approached through either manual methods or with the assistance of modern calculators. Modern calculators play an important role in problem-solving and can enhance efficiency, accuracy, and conception.

In this post, we will explore the basic definition, types, and examples of calculus.

## What is calculus?

In mathematics, calculus deals with the changes that occur in functions/things. The basic role of calculus is to find the rates of change, areas under the curve, and behavior of the functions at particular points.

Dealing with the problems in which variables of the functions change with time and some other reference variables makes calculus the backbone of mathematics. It is essentially used to deal with the properties and behavior of integral, differential, and limits.

Before discussing the types and calculations of calculus, you must have sound knowledge of functions, variables, and equations.

### Function

In mathematics, the relationship that maps elements of the domain (set of all possible input values) to codomain (range) (set of all possible output values) is said to be the function. It links every dependent variable with the independent variable that could be unique. It is generally denoted by “f(x)” or “y”

### Variable

In mathematical expression, the variable is used to represent changing quantities while in mathematical equations, the variable is used to represent unknown terms. Variables are represented by alphabets such as x, y, or z.

The variable can be dependent, independent, or constant.

- The input values that are not dependent on anyone and can be chosen freely are said to be the independent variables.
- The output values that are totally dependent on the independent variables are said to be the dependent variables.
- The fixed values that are not changed throughout the problem are said to be constant.

## Types of Calculus

The basics types of calculus are as follows:

- Limits
- Derivative
- Integral

Let’s discuss the types of calculus briefly along with the calculus problems.

### Limit Calculus

Limit is a type of calculus that is considered the heart of calculus. It deals with the properties and nature of the function at a specific point. The specific point could be finite or infinite.

In calculus, a value that a function approaches ad its independent variable gets closer and closer to a particular point is said to be the limit value. The term limit is denoted as “lim”, “Lim”, or “Lt”. Mathematically, limits are written as:

**Lt _{u→c} f(u) = L**

Where

- “u” represents the independent variable
- “c” is the value towards which “u” is approaching
- “f(u)” is the function
- “L” is the limit value.

Limit calculus is crucial for understanding the behavior of functions at particular points. It allows us to analyze rates of change (finding derivative), continuity, and the existence of vertical or horizontal asymptotes.

**Rules of Limit Calculus**

Here are a few basic rules of limit calculus that are used to find limit values from complex calculus problems.

**Sum Rule:**Lt_{u}_{→c}[f(u) + g(u)] = Lt_{u}_{→c}[f(u)] + Lt_{u}_{→c}[h(u)]**Difference Rule:**Lt_{u}_{→c}[f(u) – g(u)] = Lt_{u}_{→c}[f(u)] – Lt_{u}_{→c}[h(u)]**Product Rule**: Lt_{u}_{→c}[f(u) * g(u)] = Lt_{u}_{→c}[f(u)] * Lt_{u}_{→c}[h(u)]**Constant Rule:**Lt_{u}_{→c}[M] = M**Constant function Rule:**Lt_{u}_{→c}[M f(u)] = M Lt_{u}_{→c}[f(u)]**Power Rule:**Lt_{u}_{→c}[f(u)]^{n}= [Lt_{u}_{→c}f(u)]^{n}**Quotient Rule:**Lt_{u}_{→c}[f(u) / g(u)] = Lt_{u}_{→c}[f(u)] / Lt_{u}_{→c}[h(u)]

**How to Find Limits Problems Using Calculator and Manually?**

The limit of the function can be found easily with the help of a calculator or by using rules of limit calculus manually.

**Using a Limit calculator**

The problems of limits can be solved easily with the help of online limit calculators. You can take assistance from a limit calculator by AllMath to get the step-by-step solution in a fraction of a second.

Below are a few steps to find the limit using a calculator.

- Enter the function f(u)
- Select the independent variable
- Enter the particular point
- Hit submit button
- The step-by-step solution will come below the submit button

**Using manual method**

Here is an example of finding limits using a manual method.

Find the limit of f(u) = (3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12) if “u” approaches “1”.

**Solution **

**Step 1:** First of all, write the given function according to the general expression of the limit.

f(u) = (3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)

Lt_{u}_{→}_{c} [f(u)] = Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)]

**Step 2:** Now apply the quotient rule of limit calculus to the above expression.

Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)] = Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6)] / Lt_{u}_{→1} [ (5u^{2} + 13u – 12)]

**Step 3:** Now use the sum and difference rules of limit calculus.

Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)] = (3 Lt_{u}_{→1} [u^{2}] – 2Lt_{u}_{→1} [u^{3}] + Lt_{u}_{→1} [6]) / (5 Lt_{u}_{→1} [u^{2}] + 13 Lt_{u}_{→1} [u] – Lt_{u}_{→1} [12])

**Step 4:** Now take out the constant terms outside the notation of limit.

Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)] = (3 Lt_{u}_{→1} [u^{2}] – 2Lt_{u}_{→1} [u^{3}] + Lt_{u}_{→1} [6]) / (5 Lt_{u}_{→1} [u^{2}] + 13 Lt_{u}_{→1} [u] – Lt_{u}_{→1} [12])

**Step 5:** Now put u = 1 in the above expression to find the limit.

Lt_{u}_{→1} [(3u^{2} – 2u^{3} + 6) / (5u^{2} + 13u – 12)] = (3 [1^{2}] – 2 [1^{3}] + [6]) / (5 [1^{2}] + 13 [1] – [12])

= (3 [1] – 2 [1] + [6]) / (5 [1] + 13 [1] – [12])

= (3 – 2 + 6) / (5 + 13 – 12)

= (1 + 6) / (18 – 12)

= (7) / (6)

= 1.1667

### Integral Calculus

In calculus, the integral is the basic type that is essential for computing areas under the curve, finding accumulated quantities, and analyzing the behavior of the function. The integral is also used to find the antiderivative whose original function is derivative.

The two basic branches of integral calculus are definite integral and indefinite integral. The definite integral deals with the boundary values to find the area under the curve with the help of the fundamental theorem of calculus.

_{c}∫^{d} f(u) du = [F(u)]^{d}_{c} = F(d) – F(c)

Where

- c & d = the boundary values
- f(u) = differential function
- F(u) = new function
- u = integrating variable
- F(d) – F(c) = fundamental theorem of calculus

While the indefinite integral doesn’t deal with the boundary values and its main function is to find the new function with an arbitrary constant. The integral calculus is denoted by a symbol “∫”.

∫ f(u) du = F(u) + C

Where

- f(u) = differential function
- F(u) = new function
- u = integrating variable
- C = integrating constant

#### Rules of Integral Calculus

Here are a few general rules of integral calculus that are used to solve complex calculus problems related to finding limits.

**Sum Rule:** ʃ [f(u) + h(u)] du = ʃ [f(u)] du + ʃ [h(u)] du + C

**Difference Rule:**ʃ [f(u) – h(u)] du = ʃ [f(u)] du – ʃ [h(u)] du + C**Constant Rule:**ʃ M du = M * u + C**Constant function Rule:**ʃ [M f(u)] du = M ʃ [f(u)] du**Power Rule:**ʃ f(u)^{n}du = f(u)^{m + 1}/ (m + 1) + C

**How to Find Integral Problems Using Calculator and Manually?**

The integral of the function can be found easily with the help of a calculator or by using the rules of integral calculus manually.

**Using an integral calculator**

The problems of integration can be solved easily with the help of online integral calculators. You can take assistance from an antiderivative calculator to get the step-by-step solution in a fraction of a second.

Below are a few steps to find integral using a calculator.

- Select the type i.e., definite or indefinite
- Enter the function f(u)
- Select the independent variable
- Enter the boundary values in case of the definite integral
- Hit calculate button
- The step-by-step solution will come below the calculate button

**Using manual method**

Here is an example of finding integral using the manual method.

**Example 1**

Find the new function if h(u) = 12u^{3 }+ 15u^{2} – 9u^{4} – 4cos(u) + 4u^{3} with respect to “u”

**Solution **

**Step 1:** First of all, take the given function and apply the notation of antiderivative to that function.

h(u) = 12u^{3 }+ 15u^{2} – 9u^{4} – 4cos(u) + 4u^{3}

ʃ h(u) du = ʃ [12u^{3 }+ 15u^{2} – 9u^{4} – 4cos(u) + 4u^{3}] du

**Step 2:** Now use the addition and subtraction rules of integral calculus to the above expression to apply the integral notation to each term of the function.

ʃ [12u^{3 }+ 15u^{2} – 9u^{4} – 4scos(u) + 4u^{3}] du = ʃ [12u^{3}] du + ʃ [15u^{2}] du – ʃ [9u^{4}] du – ʃ [4cos(u)] du + ʃ [4u^{3}] du

**Step 3:** Now take out the constant term outside the notation of integral.

ʃ [12u^{3 }+ 15u^{2} – 9u^{4} – 4scos(u) + 4u^{3}] du = 12ʃ [u^{3}] du + 15ʃ [u^{2}] du – 9ʃ [u^{4}] du – 4ʃ [cos(u)] du + 4ʃ [u^{3}] du

**Step 4: **Now integrate the above expression with the help of power and trigonometry properties.

ʃ [12u^{3 }+ 15u^{2} – 9u^{4} – 4scos(u) + 4u^{3}] du = 12 [u^{3 + 1 }/ 3 + 1] + 15 [u^{2 + 1 }/ 2 + 1] – 9 [u^{4 + 1 }/ 4 + 1] – 4 [sin(u)] + 4 [u^{3 + 1 }/ 3 + 1] + C

ʃ [12u^{3 }+ 15u^{2} – 9u^{4} – 4scos(u) + 4u^{3}] du = 12 [u^{4 }/ 4] + 15 [u^{3 }/ 3] – 9 [u^{5 }/ 5] – 4 [sin(u)] + 4 [u^{4 }/ 4] + C

= 12/4 [u^{4}] + 15/3 [u^{3}] – 9/5 [u^{5}] – 4 [sin(u)] + 4/4 [u^{4}] + C

= 3 [u^{4}] + 5 [u^{3}] – 9/5 [u^{5}] – 4 [sin(u)] + 1 [u^{4}] + C

= 3u^{4 }+ 5u^{3} – 9u^{5}/5 – 4sin(u) + u^{4} + C

= 4u^{4 }+ 5u^{3} – 9u^{5}/5 – 4sin(u) + C

## Derivative Calculus

In calculus, the derivative is a fundamental concept that is used to deal with the instantaneous rate of change and finding the slope of the tangent line. The change in the function can be determined with the help of this type of calculus.

Geometrically, it represents the slope of the tangent line to the curve at that point. The notations of derivative calculus are:

- Leibniz’s notation = dy/dx
- Newton’s notation = f’
- prime notation = f'(x)

The repetition in taking the derivative of a similar function makes a higher order derivative. Such as second derivative (d^{2}y/dx^{2}), third derivative(d^{3}y/dx^{3}), etc. The behavior and characteristics of functions can only be understood by having a solid understanding of derivative calculus.

#### Rules of Derivative Calculus

Here are a few general rules of derivatives that are used to solve complex calculus problems related to finding differentials.

**Sum Rule:**d/du [f(u) + h(u)] = d/du [f(u)] + d/du [h(u)]**Difference Rule:**d/du [f(u) – h(u)] = d/du [f(u)] – d/du [h(u)]**Product Rule**: d/du [f(u) * h(u)] = h(u) d/du [f(u)] + f(u) d/du [h(u)]**Constant Rule:**d/du [M] = 0**Constant function Rule:**d/du [M f(u)] = M d/du [f(u)]**Power Rule:**d/du [f(u)]^{n}= n [f(u)]^{n-1}d/du [f(u)]**Quotient Rule:**d/du [f(u) / h(u)] = 1/[h(u)]^{2}[h(u) d/du (f(u)) – f(u) d/du (h(u))]

**How to Find Derivative Problems Using Calculator and Manually?**

The derivative of the function can be found easily with the help of a calculator or by using the rules of derivative calculus manually.

**Using a derivative calculator**

The problems of differential can be solved easily with the help of online derivative calculators. You can take assistance from a derivative calculator by MeraCalculator to get the step-by-step solution in a fraction of a second.

Below are a few steps to find derivatives using a calculator.

- Enter the function f(u)
- Select the independent variable
- Enter the order of derivative
- Hit submit button
- The step-by-step solution will come below the submit button

**Using manual method**

Here is an example of finding derivatives using the manual method.

Evaluate the differential of “f(u) = 16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4}) with respect to “u”.

**Solution**

**Step 1:** First of all, take the given function and write the notation of the derivative to it.

f(u) = 16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})

d/du [f(u)] = d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})]

**Step 2:** Now use the addition and subtraction rules of derivative calculus to the above expression to apply the derivative notation to each term of the function.

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = d/du [16u^{2}] + d/du [2u^{3}] – d/du [12sin(u)] + d/du [ (u^{3} / u^{4})]

**Step 3:** Now apply the quotient rule to the above expression

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = d/du [16u^{2}] + d/du [2u^{3}] – d/du [12sin(u)] + [1/(u^{4})^{2} [u^{4} d/du (u^{3}) – u^{3} d/du (u^{4})]]

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = d/du [16u^{2}] + d/du [2u^{3}] – d/du [12sin(u)] + u^{4}/(u^{8}) d/du (u^{3}) – u^{3}/u^{8} d/du (u^{4})]]

**Step 4:** Now take the constant coefficient outside the notation.

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = 16d/du [u^{2}] + 2d/du [u^{3}] – 12d/du [sin(u)] + 1/u^{4} d/du (u^{3}) – 1/u^{5} d/du (u^{4})]]

**Step 5:** Now use the power rule to differentiate the above expression.

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = 16 [2 u^{2-1}] + 2 [3 u^{3-1}] – 12 [cos(u)] + 1/u^{4} (3 u^{3-1}) – 1/u^{5} (4 u^{4-1})]

d/du [16u^{2} + 2u^{3} – 12sin(u) + (u^{3} / u^{4})] = 16 [2 u^{1}] + 2 [3 u^{2}] – 12 [cos(u)] + 1/u^{4} (3 u^{2}) – 1/u^{5} (4 u^{3})]

= 16 [2 u] + 2 [3 u^{2}] – 12 [cos(u)] + 1/u^{4} (3 u^{2}) – 1/u^{5} (4 u^{3})]

= 32u + 6 u^{2} – 12cos(u) + 3u^{2}/u^{4} – 4u^{3}/u^{5}

= 32u + 6u^{2} – 12cos(u) + 3/u^{2} – 4/u^{2}

= 32u + 6u^{2} – 12cos(u) + 3u^{-2} – 4u^{-2}

= 32u + 6u^{2} – 12cos(u) – u^{-2}

## Final Words

Calculus is the fundamental branch of mathematics that deals with the properties and calculations of limit, derivative, and integral. The problems of calculus can be solved easily with the help of online tools and manual methods.