Calculus is a rather tough topic in mathematics as it involves complex kinds of problems. Students face many difficulties in understanding this concept. There are two main kinds of calculus i.e., integration and differentiation.
Both types play a vital role in geometry for finding various concepts. In this post, we are going to discuss integration along with definitions, formulas, examples, and solutions.
What is integration?
In calculus, the process of finding the integral of the function with respect to the integrating variable is known as integration. It is widely used in geometry to find the area under the curve along with a graph.
It finds the new function and the numerical values of the function with or without using the boundary values. The integrating variable is very essential for finding the integral of the function.
The integral is of two types that are used to integrate the functions with or without using upper and lower limits. These types are
- Definite integral
- Indefinite integral
Let us briefly describe the types of integration with their formulas.
Definite integral
To find the numerical values of the function the definite integral is used. It integrates the function with respect to the integrating variable and after that, it uses the upper and lower limit values.
In simple words, the definite integral finds the numerical values of the function with the help of boundary values. The limit values are applied to the function by using a fundamental theorem of calculus. According to this theorem,
The upper limit is applied to the integral function and then the lower limit with a difference sign among them. Below is a general expression for the definite integral.
Where x and y are the boundary values, f(w) is the integrand, w is the integrating variable, F(v) – F(u) is the theorem of calculus, and M is the final result after applying the fundamental theorem.
Indefinite integral
To find a new function or to calculate the reverse of a derivative this type of integration is used. The indefinite integral is calculating the integral of the function by using the variable of integration. In this type of integration, the constant of integration is used.
Below is a general expression for the indefinite integral.
ʃ f(w) dw = F(w) + C
where f(w) is the integrand, w is the integrating variable, F(w) is a new function, and C is the integral constant.
How to calculate integration problems?
To solve the problems of integration, there are two methods.
- By using an integral calculator
- Manually
Have a look at both the methods of finding integral calculus problems.
By using an integral calculator
There are hundreds of online tools present online as it is the era of technology. The online tools are created for helping students who face difficulties while solving complex problems and completing assignments.
The integration problems can be solved easily by using an integral calculator.
How to use this calculator?
Follow the below steps.
- Select the definite or indefinite integral.
- Enter the function into the required input field.
- Select the integrating variable “x” is selected by default.
- Enter the number of upper and lower limits in the case of the definite integral.
- Fill the captcha
- Press the calculate button.
- The solution with steps will come below the calculate button.
Manually
Example 1: For indefinite integral
Calculate the indefinite integral of the given function with respect to “w”.
f(w) = 12w4 + 3w2 – 7w6 + 5cos(w) + 12w
Solution
Step I: Apply the notation of indefinite integral to the given function.
ʃ f(w) dw = ʃ [12w4 + 3w2 – 7w6 + 5cos(w) + 12w] dw
Step II: Now apply the notation of indefinite integral separately to each function with the help of sum and difference rules.
ʃ [12w4 + 3w2 – 7w6 + 5cos(w) + 12w] dw = ʃ [12w4] dw + ʃ [3w2] dw – ʃ [7w6] dw + ʃ [5cos(w)] dw + ʃ [12w] dw
Step III: Use the constant function rule and take the constant coefficients outside the integral notation.
= 12ʃ [w4] dw + 3ʃ [w2] dw – 7ʃ [w6] dw + 5ʃ [cos(w)] dw + 12ʃ [w] dw
Step IV: Integrate the above expression.
= 12 [w4+1 / 4 + 1] + 3 [w2+1 / 2 + 1] – 7 [w6+1 / 6 + 1] + 5 [sin(w)] + 12 [w1+1 / 1 + 1] + C
= 12 [w5 / 5] + 3 [w3 / 3] – 7 [w7 / 7] + 5 [sin(w)] + 12 [w2 / 2] + C
= 12/5 [w5] + 3/3 [w3] – 7/7 [w7] + 5 [sin(w)] + 12/2 [w2] + C
= 12/5 [w5] + [w3] – [w7] + 5 [sin(w)] + 6 [w2] + C
= 12w5/5 + w3 – w7 + 5sin(w) + 6w2 + C
Example 2: For definite integral
Calculate the definite integral of the given function with respect to “w”.
f(t) = 2t5 – 12t3 + 21t2 + 3sin(t) in the interval of [3, 4].
Solution
Step I: Apply the notation of indefinite integral to the given function.
ʃf(t) dt = ʃ[2t5 – 12t3 + 21t2 + 3sin(t)] dt
Step II: Now apply the notation of indefinite integral separately to each function with the help of sum and difference rules.
[2t5 – 12t3 + 21t2 + 3sin(t)] dt = [2t5] dt – [12t3] dt + [21t2] dt + [3sin(t)] dt
Step III: Use the constant function rule and take the constant coefficients outside the integral notation.
= 2[t5] dt – 12[t3] dt + 21[t2] dt + 3[sin(t)] dt
Step IV: Now integrate the above expression.
= 2 [t5+1 / 5 + 1]43 – 12 [t3+1 / 3 + 1]43 + 21 [t2+1 / 2 + 1]43 + 3 [-cos(t) ]43
= 2 [t6 / 6]43 – 12 [t4 / 4]43 + 21 [t3 / 3]43 + 3 [-cos(t) ]43
= 2/6 [t6]43 – 12/4 [t4]43 + 21/3 [t3]43 + 3 [-cos(t) ]43
= 1/3 [t6]43 – 3 [t4]43 + 7 [t3]43 + 3 [-cos(t) ]43
= 1/3 [t6]43 – 3 [t4]43 + 7 [t3]43 – 3 [cos(t) ]43
Step 4: Apply the boundary values.
= 1/3 [46 – 36] – 3 [44 – 34] + 7 [43 – 33] – 3 [cos(4) – cos(3)]
= 1/3 [4096 – 729] – 3 [256 – 81] + 7 [64 – 27] – 3 [cos(4) – cos(3)]
= 1/3 [3367] – 3 [175] + 7 [37] – 3 [cos(4) – cos(3)]
= 3367/3 – 525 + 259 – 3 [cos(4) – cos(3)]
= 1122.33 – 525 + 259 – 3 [cos(4) – cos(3)]
= 597.33 + 259 – 3 [cos(4) – cos(3)]
= 856.33 – 3 [cos(4) – cos(3)]
Conclusion
Now the difficulty of solving complex integration problems can be reduced by using the methods of solving the problems. Now you can grab all the basics of integration from this post such as definition, types, formulas, and solved examples
