Integral Calculus: Explanation, Types, Properties, and Examples
In mathematics, the integral is one of the
essential types of calculus that is frequently used to find the area under the
curve and antiderivative of the function. This branch of calculus is frequently
used to find the volume and other quantities that are difficult to evaluate by
other methods.
It is an essential type of calculus along
with differential calculus that deals with the instantaneous rate of change. In
this article, we will discuss the term integral calculus along with its
properties and types with solved examples.
What is integral in calculus?
In mathematics, calculus is frequently used
for dealing with change and motions. There are two well-known branches of calculus
such as integral calculus and differential calculus. The integral calculus is
used to find the antiderivative, area under the curve, and volume.
While the other type of calculus is used to
calculate the rate of change of the function with respect to the independent
variable. The integral is the most frequently used type of calculus that is
essential for the calculations of various things to get rid of difficult ways.
Types of Integral Calculus
There are two types of integral calculus
such as antiderivative and definite integral. Here is a brief introduction to
the types of the integral calculus.
Antiderivative
The antiderivative
is the branch of integral calculus that is essential to find the new function
whose original function is a differential function with respect to integrating
variables. The constant of integration will be applied with the integral of the
function.
There are no boundary values involved in
this type of integral calculus. The term antiderivative is also known as
indefinite integral. The expression of antiderivative is:
∫ f(t) dt =
F(t) + C
Where,
·
f(t) = differential function
·
F(t) = new function
·
t = integrating variable
·
C = integrating constant
Definite integral
The other type of integral calculus is
definite integral which is used to find the area and volume under the curve
with the help of boundary values of the function. The boundary values of the
function are known as the upper and lower limit values.
The boundary values are to be applied to
the new function with the help of the fundamental theorem of calculus. The expression
of the definite integral is:
_{u}∫^{v} f(t) dt = [F(t)]^{v}_{u}
= F(v) – F(u)
Where,
·
u & v = upper and lower
limit values
·
f(t) = differential function
·
F(t) = new function
·
t = integrating variable
Properties of Integral Calculus
There are several properties of integral
calculus. Here we are going to discuss some of them.
Properties |
Expressions |
Power Property |
Êƒ f(t)^{ n} dt = f(t)^{ n +
1}/ (n + 1) + C |
Constant Property |
Êƒ K dt = K * t + C |
Constant Function Property |
Êƒ K * f(t) dt = K * Êƒ f(t) dt + C |
Sum Property |
Êƒ [f(t) + g(t)] dt = Êƒ [f(t)] dt + Êƒ
[g(t)] dt + C |
Difference Property |
Êƒ [f(t) - g(t)] dt = Êƒ [f(t)] dt - Êƒ
[g(t)] dt + C |
Trigonometry Property |
Êƒ sin(t) dt = -cos(t) + C Êƒ cos(t) dt = sin(t) + C Êƒ tan(t) dt = ln |sec(t)| + C |
Exponential Property |
Êƒ e^{t} dt = e^{t} + C |
Product Property |
Êƒ f(t) * g’(t) dt = f(t) g(t) - Êƒ g(t)
* f’(t) dt |
How to calculate integral calculus problems?
There are two ways of calculating integral
calculus problems such as solving them with the help of an online
integral calculator and the other is solving them manually with the
help of properties and formulas.
Example 1
Calculate the antiderivative of g(t) = 12t^{2
}+ 2cos(t) – 4t^{3} + 12t – 100 with respect to “t”
Solution
Using manual method
Step 1: Write the given
differential function and apply the notation of antiderivative to it.
Êƒ g(t) dt = Êƒ [12t^{2 }+ 2cos(t) – 4t^{3} + 12t – 100] dt
Step 2: Now apply the sum and
difference properties of integral calculus to the above expression.
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = Êƒ [12t^{2}] dt^{ }+ Êƒ [2cos(t)] dt – Êƒ [4t^{3}]
dt + Êƒ [12t] dt – Êƒ [100] dt
Step 3: Now use the constant function property of integral calculus.
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 12Êƒ [t^{2}] dt^{ }+ 2Êƒ [cos(t)] dt – 4Êƒ [t^{3}]
dt + 12Êƒ [t] dt – Êƒ [100] dt
Step 4: Now integrate the above expression with the help of power and
trigonometry properties.
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 12 [t^{2 + 1}_{ }/ 2 + 1]^{ }+ 2 [sin(t)] – 4
[t^{3 + 1}_{ }/ 3 + 1] + 12 [t^{1 + 1}_{ }/ 1 +
1] – [100 t] + C
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 12 [t^{3}_{ }/ 3]^{ }+ 2 [sin(t)] – 4 [t^{4}_{
}/ 4] + 12 [t^{2}_{ }/ 2] – 100 [t] + C
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 12/3 [t^{3}]^{ }+ 2 [sin(t)] – 4/4 [t^{4}] + 12/2
[t^{2}] – 100 [t] + C
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 4 [t^{3}]^{ }+ 2 [sin(t)] – 1 [t^{4}] + 6 [t^{2}]
– 100 [t] + C
Êƒ [12t^{2 }+
2cos(t) – 4t^{3} + 12t – 100] dt = 4t^{3 }+ 2sin(t) – t^{4}
+ 6t^{2} – 100t + C
Using calculator the result will be:
Example 2
Calculate the area under the curve between
the interval of [2, 3] with respect to “v”
Example 1
Calculate the numerical value of g(t) = 2t^{3
}+ 6t^{2} – 12t^{5} + 16t – 4t^{4} with respect to
“t” in the interval of [2, 3]
Solution
Step 1: Write the given
differential function and apply the notation of integral to it with the
interval.
_{u}Êƒ^{v} g(t) dt = _{2}Êƒ^{3}
[2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}] dt
Step 2: Now apply the sum and
difference properties of integral calculus to the above expression.
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = _{2}Êƒ^{3} [2t^{3}]
dt^{ }+ _{2}Êƒ^{3}
[6t^{2}] dt – _{2}Êƒ^{3}
[12t^{5}] dt + _{2}Êƒ^{3}
[16t] dt – _{2}Êƒ^{3}
[4t^{4}] dt
Step 3: Now use the constant function property of integral calculus.
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 2_{2}Êƒ^{3} [t^{3}]
dt^{ }+ 6_{2}Êƒ^{3}
[t^{2}] dt – 12_{2}Êƒ^{3}
[t^{5}] dt + 16_{2}Êƒ^{3}
[t] dt – 4_{2}Êƒ^{3} [t^{4}]
dt
Step 4: Now integrate the above expression with the help of power and
trigonometry properties.
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 2 [t^{3 + 1} / 3 + 1]^{3}_{2}^{ }+ 6 [t^{2
+ 1} / 2 + 1]^{3}_{2} – 12 [t^{5 + 1} / 5 + 1]^{3}_{2}
+ 16 [t^{1 + 1} / 1 + 1]^{3}_{2} – 4 [t^{4 + 1}
/ 4 + 1]^{3}_{2}
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 2 [t^{4} / 4]^{3}_{2}^{ }+ 6 [t^{3}
/ 3]^{3}_{2} – 12 [t^{6} / 6]^{3}_{2} +
16 [t^{2} / 2]^{3}_{2} – 4 [t^{5} / 5]^{3}_{2}
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 2/4 [t^{4}]^{3}_{2}^{ }+ 6/3 [t^{3}]^{3}_{2}
– 12/6 [t^{6}]^{3}_{2} + 16/2 [t^{2}]^{3}_{2}
– 4/5 [t^{5}]^{3}_{2}
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 1/2 [t^{4}]^{3}_{2}^{ }+ 2 [t^{3}]^{3}_{2}
– 2 [t^{6}]^{3}_{2} + 8 [t^{2}]^{3}_{2}
– 4/5 [t^{5}]^{3}_{2}
Step 5: Now use the
fundamental theorem of calculus to apply the upper and lower limits of the integral.
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 1/2 [3^{4} – 2^{4}]^{ }+ 2 [3^{3} – 2^{3}]
– 2 [3^{6} – 2^{6}] + 8 [3^{2} – 2^{2}] – 4/5 [3^{5} – 2^{5}]
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 1/2 [81 – 16]^{ }+ 2 [27 – 8] – 2 [729 – 64] + 8 [9 – 4] – 4/5 [243 – 32]
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 1/2 [65]^{ }+ 2 [19] – 2 [665] + 8 [5]
– 4/5 [211]
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 165/2^{ }+ 38 – 1330 + 40 –
844/5
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 82.5^{ }+ 38 – 1330 + 40 –
844/5
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 82.5^{ }+ 38 – 1330 + 40 –
168.8
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = 120.5 – 1330 + 40 – 168.8
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = -1209.5 + 40 – 168.8
_{2}Êƒ^{3} [2t^{3 }+ 6t^{2} – 12t^{5} + 16t – 4t^{4}]
dt = -1338.3
Final Words
The integral is the fundamental branch of
calculus that helps you to find the antiderivative, the area under the curve,
and the volume. The types of the integral are used to find the integral with or
without limits. The problems above will help to solve the integrals manually.
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