# What is Integral Calculus & How to Evaluate It Manually and Using Calculator?

The term calculus is one of the fundamental branches of mathematics that deals with the properties and calculations of change and motion. This mathematical branch is subdivided into two main components known as derivative and integral.

The term derivative in calculus deals with the rate of change of functions respective to the independent variable. The slope of the tangent line can be evaluated with the help of this branch of calculus.

The other type of calculus named integral calculus inverted the results that derivative done and finding the areas under the curve. In this article, we will explore integral in calculus along with its types, and evaluating process manually and using calculator.

**What is integral in Calculus?**

An integral in calculus is a fundamental concept that helps to find the area under the curve by relying on the given interval of the function. This branch of calculus is mainly used to find the new function whose original function is derivative.

In calculus, the integral of a function f(z) will be represented as ∫ f(z) dz, where the integrating variable “dz” infinitesimal element of the variable “z”. There are further two types of integral in calculus such as definite integral and indefinite integral. *copyright©iasexpress.net*

**Definite Integral **

In calculus, the definite integral is a basic concept that deals with the numerical value of the given function over a given interval. The process of finding the numeric value of the function over the given interval is said to be finding the area under the curve.

The fundamental theorem of calculus plays a vital role in the calculation of areas under the curves. This theorem is helpful for applying the upper and lower limit values to the integrated function to evaluate the numeric value.

The formula used to solve the problems of this kind of integral is:

_{p}**∫**^{q}** f(z) dz = |F(z)|**_{p}^{q}** = F(q) – F(p)**

where, p & q are the interval points of the given function, “z” is the independent variable of the function, f(z) is the given differential function, F(z) is the integrated function, & F(q) – F(p) denotes the fundamental theorem of calculus.

**Indefinite Integral**

In calculus, the indefinite integral is used to find the new function whose genuine function is differential. This branch of calculus is also known as primitive function & antiderivative. The name antiderivative shows that this branch of integral works as the inverse of differential. *copyright©iasexpress.net*

There is no interval involved in this type of integral. The constant of integration is written along with the integrated function. The formula used to solve the problems of this kind of integral is:

**∫ f(z) dz = F(z) + C**

where, “z” is the independent variable of the function, f(z) is the given differential function, F(z) is the integrated function, & C is the constant of integration.

**Rules of Integral in Calculus**

There are several rules of integral calculus for dealing with various kind of problems.

**Sum Rule:**ʃ [f(z) + h(z)] dz = ʃ [f(z)] dz + ʃ [h(z)] dz + C**Difference Rule:**ʃ [f(z) – h(z)] dz = ʃ [f(z)] dz – ʃ [h(z)] dz + C**Constant Rule:**ʃ K dz = K * z + C**Constant function Rule:**ʃ [K h(z)] dz = K ʃ [h(z)] dz**Power Rule:**ʃ f(z)^{m}dz = f(z)^{m + 1}/ (m + 1) + C

**How to evaluate the integral of the function manually?**

There are various methods for solving integrals manually.

**Direct Integration Method**

In calculus, the direct integration method is the general way to integrate the given function with the help of rules of integration. This method is applicable to simple functions that can be evaluated with the help of sum, difference, power, and constant function rules. *copyright©iasexpress.net*

**Example**

Find the antiderivative of the given function with respect to “z”.

h(z) = 2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}

**Solution **

**Step 1:** Write the given function according to the mathematical representation of the antiderivative.

h(z) = 2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}

ʃ h(z) dz = ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz

**Step 2:** Now use the rule of sum and difference to make the function simple.

ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz = ʃ [2z^{3}] dz^{ }– ʃ [15z^{2}] dz + ʃ [6z^{5}] dz – ʃ [5cos(z)] dz + ʃ [15z^{2}] dz** **

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

ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz = 2ʃ [z^{3}] dz^{ }– 15ʃ [z^{2}] dz + 6ʃ [z^{5}] dz – 5ʃ [cos(z)] dz + 15ʃ [z^{2}] dz** **

**Step 4: **Now by direct integration method, take the power rule of the above expression to find the antiderivative.

ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz = 2 [z^{3 + 1 }/ 3 + 1]^{ }– 15 [z^{2 + 1 }/ 2 + 1] + 6 [z^{5 + 1 }/ 5 + 1] – 5 [sin(z)] + 15 [z^{2 + 1 }/ 2 + 1] + C

ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz = 2 [z^{4 }/ 4]^{ }– 15 [z^{3 }/ 3] + 6 [z^{6 }/ 6] – 5 [sin(z)] + 15 [z^{3 }/ 3] + C

ʃ [2z^{3 }– 15z^{2} + 6z^{5} – 5cos(z) + 15z^{2}] dz = 2/4 [z^{4}]^{ }– 15/3 [z^{3}] + 6/6 [z^{6}] – 5 [sin(z)] + 15/3 [z^{3}] + C

= 1/2 [z^{4}]^{ }– 5 [z^{3}] + 1 [z^{6}] – 5 [sin(z)] + 5 [z^{3}] + C

= z^{4}/2^{ }– 5z^{3} + z^{6} – 5sin(z) + 5z^{3} + C

= z^{4}/2 + z^{6} – 5sin(z) + C

**Integration by Substitution**

In calculus, integration by substitution is another method for finding the integral of the given function. this method is also known as U-substitution as the variable “u” is substituted to the given function to replace some terms to make that function simple. *copyright©iasexpress.net*

This method makes the process of integration easier for the complex function as the system of u & du will be substituted in place of complex parts.

**Example**

Find the antiderivative of the given function with respect to “z”.

h(z) = z^{4 }* e^{z^5}

**Solution **

**Step 1:** Write the given function according to the mathematical representation of the antiderivative.

h(z) = z^{4 }* e^{z^5}

ʃ h(z) dz = ʃ [z^{4 }* e^{z^5}] dz

**Step 2:** Find the term that you want to substitute.

We’ll substitute u = z^{5}

As the differential of z^{5} will give us 5z^{4} which is the part of the given function.

**Step 3:** Now find the integrating variable “du”.

As u = z^{5}

Then, du/dz = 5z^{5-1}

du/dz = 5z^{4}

du = 5z^{4} dz

du/5 = z^{4} dz

**Step 4:** Now substitute the value of u & du to the integral function.

ʃ [z^{4 }* e^{z^5}] dz = ʃ e^{u} /5 du

**Step 5:** Now integrate the above exponential function.

ʃ e^{u} /5 du = 1/5 ʃ e^{u} du

As the integral of the exponential function remains the same

ʃ e^{u} /5 du = 1/5 e^{u} + C

**Step 6:** Now place the value of u to the above expression.

ʃ [z^{4 }* e^{z^5}] dz = 1/5 e^{z^5} + C

**Integration by Parts**

In calculus, integration by parts is the basic method for finding the integral of the function. it deals with the product of two functions. This method is derived from the product rule of derivative. There are two terms involved in this function such as u & dv. *copyright©iasexpress.net*

The general form of this method is:

∫u * v = u * ∫ v dv – ∫ du/dz [∫v dv] dv

This method is applicable for dealing with the product of two functions, exponential function, logarithmic functions, etc.

**Example**

Find the antiderivative of the given function with respect to “z”.

h(z) = z ln(z)

**Solution **

**Step 1:** Write the given function according to the mathematical representation of the antiderivative.

h(z) = z ln(z)

ʃ h(z) dz = ʃ [z ln(z)] dz

**Step 2:** Identify u & v

u = ln(z)

v = z

**Step 3:** Take the formula of substitution by parts

∫u * v = u * ∫ v dv – ∫ du/dz [∫v dv] dv

**Step 4:** Now substitute the values to the above formula and simplify it.

ʃ [z ln(z)] dz = ln(z) * ∫ z dz – ∫ d/dz (ln(z)) [∫ z dz] dz

ʃ [z ln(z)] dz = ln(z) * [z^{1 + 1} / 1 + 1] – ∫ d/dz (ln(z)) [z^{1 + 1} / 1 + 1] dz

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ∫ (1/z) [z^{2} / 2] dz

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ∫ [z^{2} / 2z] dz

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ∫ [z / 2] dz

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ½ ∫ [z] dz

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ½ [z^{1 + 1} / 1 + 1]

ʃ [z ln(z)] dz = ln(z) * [z^{2} / 2] – ½ [z^{2} / 2] *copyright©iasexpress.net*

ʃ [z ln(z)] dz = z^{2} ln(z)/2 – z^{2}/4

**Step 5:** Add the constant of integration.

ʃ [z ln(z)] dz = z^{2} ln(z)/2 – z^{2}/4 + C

**How to Evaluate Integral Using a Calculator?**

There are hundreds of online calculators available to find the numerical problems of integral calculus. Many tools provide only results of the given function that is not sufficient for understanding and seeking guidance for evaluating limit.

Although some of them provide solutions with steps such as an integral calculator by MeraCalculator. This calculator will allow you to solve integral problems with the help of definite and indefinite integrals with steps.

**Example**

Find the antiderivative of the given function with respect to “z”.

h(z) = z ln(z)

**Solution**

**Step 1:** Write the function into the input box.

**Step 2:** Select the variable “z”.

**Step 3:** Hit the submit button.

**Step 4:** Result

**Wrap Up**

Now you can grab all the basics of evaluating integrals manually and using a calculator from this post. We have discussed definition, types, rules, and methods of evaluating integral with the help of solved examples.

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