In mathematics, the integral is one of the two main branches of calculus. This branch of calculus is used to find the area under the curve. It is usually used to find the integral of the functions with respect to integrating variables.
While the other type of calculus is used to find the differential of the function withrespect to the independent variable. Both types of calculus can be defined easily by using limit calculus. In this article, we will learn the basics and how to calculate the problems of integral calculus.
What Is Integral Calculus?
In mathematics, a new function whose original function is derivative or the numerical value of the function to determine the area under the curve with the help of upper and lower limit values is known as integral calculus.
It is denoted by the “ʃ” symbol. Integral is the reverse process of derivatives. In simple words, integral reverse the work that a differential does.
Types of integral
There are two general types of integral.
- Definite integral
- Indefinite integral
Let us learn a brief introduction to both kinds of integral.
1. Definite integral
In calculus, the definite integral is used to find the numerical value of the function by using the upper limit and lower limit of the function. The numerical value of the integral is helpful in finding the area under the curve.
It is denoted by “”
The general expression of this type of integral is:
h(u) du = H(y) – H(x) = N
- Where x & y are the upper and lower limit values of the function respectively.
- h(u) is the given integrand function.
- “u” is the variable of integration.
- H(y) – H(x)is the integral function after applying the upper and lower limit value with the help of the fundamental theorem of calculus.
- After calculating the fundamental theorem, N is the final result.
2. Indefinite integral
In calculus, the indefinite integral is used to find a new function whose original function is derivative. The new function is used to denote the integral of the function. It is denoted by the “ʃ” symbol.
ʃ h(u) du = H(u) + C
- h(u) is the given integrand function.
- “u” is the variable of integration.
- H(u) is the new function after integration.
- C is the constant of integration.
Rules of Integral Calculus
Here are some commonly used rules of integral in calculus.
| Rule name | Rules |
| Difference rule | ʃ [h(u) – g(u)] du =ʃ [h(u)]du – ʃ [g(u)]du |
| constant rule | ʃ k du = k * u + c |
| Sum rule | ʃ [h(u) + g(u)] du =ʃ [h(u)]du + ʃ [g(u)]du |
| constant function rule | ʃ (k * h(u) du = k ʃ (h(u) du |
| power rule | ʃ [(h(u)]n du = [h(u)]n+1 / n + 1] |
| Exponential rule | ʃ eu du = eu +C |
| Reciprocal rule | ʃ (1/u) du = ln|u| + C |
How To Calculate The Problems Of Integral?
By using the rules and general expression of the definite and indefinite integral, problems of integral can be solved easily.
Example 1: For indefinite integral
Calculate the indefinite integral of h(u) = 4u3 + 3u – 5u4 + 9cos(u) + 7u + 15 with respect to u.
Solution
Step I:Take the given function h(u) and write the integral notation with the function.
h(u) = 4u3 + 3u – 5u4 + 9cos(u) + 7u + 15
integrating variable = u
ʃh(u) du = ʃ [4u3 + 3u – 5u4 + 9cos(u) + 7u + 15] du
Step II:Now use the sum and difference rule of integral and write the integral notation to each function separately and write the constant coefficients outside the function.
ʃ [4u3 + 3u – 5u4 + 9cos(u) + 7u + 15] du = ʃ [4u3] du+ ʃ [3u] du – ʃ [5u4] du + ʃ [9cos(u)] du + ʃ [7u] du + ʃ [15] du
= 4 ʃ [u3] du+ 10ʃ [u] du – 5 ʃ [u4] du + 9 ʃ [cos(u)] du + ʃ [15] du
Step III:Integrate the above expression by using the trigonometric and power rules of integral.
= 4 [u3+1 / 3 + 1] + 10 [u1+1 / 1 + 1] – 5[u4+1 / 4 + 1] + 9 [sin(u)]+ [15u]+ C
= 4 [u4 / 4] + 10 [u2 / 2] – 5 [u5 / 5] + 9 [sin(u)] + [15u] + C
= 4/4 [u4] + 10/2 [u2] – 5/5 [u5] + 9 [sin(u)] + [15u] + C
= [u4] + 5 [u2] – [u5] + 9 [sin(u)] + [15u] + C
= u4 + 5u2 – u5 + 9sin(u) + 15u + C
To get rid of these long calculations use an integral calculator with steps.
Example 2: For definite integral
Calculate the definite integral of h(u) = 3u5 – 4s2 + 5s4 + 2s + 5 with respect to u and the upper and lower limits are (3, 5)
Solution
Step I:Take the given function h(u) and write the integral notation with the function.
h(u) = 3u5 – 4s2 + 5s4 + 2s + 5
[h(u)] du =
[3u5 – 4s2 + 5s4 + 2s + 5] du
Step II:Now use the sum and difference rule of integral and write the integral notation to each function separately and write the constant coefficients outside the function.
[3u5 – 4s2 + 5s4 + 2s + 5] du =
[3u5] du –
[4s2] du +
[5s4] du +
[2s] du +
[5] du
= 3[u5] du – 4
[s2] du + 5
[s4] du + 2
[s] du +
[5] du
Step 4:Integrate the above expression by using the power rule of integral.
= 3 [u5+1 / 5 + 1]53 – 4 [s2+1 / 2 + 1]53+ 5 [s4+1 / 4 + 1]53+ 2 [s1+1 / 1 + 1]53+ [5u]53
= 3 [u6 / 6]53 – 4 [s3 / 3]53 + 5 [s5 / 5]53 + 2 [s2 / 2]53 + [5u]53
= 3/6 [u6]53 – 4/3 [s3]53 + 5/5 [s5]53 + 2/2 [s2]53 + 5 [u]53
= 1/2 [u6]53 – 4/3 [s3]53+ [s5]53+ [s2]53 + 5 [u]53
Step 5:Now apply the boundary values with the help of the fundamental theorem of calculus.
= 1/2 [56 – 36] – 4/3 [53 – 33] + [55 – 35] + [52 – 32] + 5 [5 – 3]
= 1/2 [15625 – 729] – 4/3 [125 – 27] + [3125 – 243] + [25 – 9] + 5 [5 – 3]
= 1/2 [14896] – 4/3 [98] + [2882] + [16] + 5 [2]
= 7448 – 392/3+2882 +16 + 10
= 7448 – 130.67+2882 +16 + 10
= 7317.33+2882 +16 + 10
= 10199.33 +16 + 10
= 10215.33 + 10
= 10225.33
Summary
In this article, we have covered all the basics of integral and how to calculate its problems along with a lot of examples. Now you can grab all the basics of definite and indefinite integral easily just by learning the above post.