In mathematics, calculus is a conceptual branch as compared to other branches. It is widely used in various branches of science. There are further subtypes of this kind of mathematics. Such as continuity, limit, integral, differential, etc.
In this post, we are going to discuss integral calculus deeply with explanations and calculations.
In calculus, a term that is used to evaluate the area under the curve is known as the integral. It is usually used to find a new function whose original function is the differential or numerical value of the function by using the boundary points.
The boundary values are involved in the interval points like upper and lower points. The integrating variable is essential for finding the integral of the function. This branch of calculus has two types
An integral calculator with steps can be used to calculate the definite or indefinite integral. Here we are going to discuss the kinds of integral along with explanations, formulas, and solved examples.
The first kind of integration that is helpful for finding the numerical value of the function is known as the definite integral. It is frequently used in integral calculus and using the upper and lower limit values of the function.
The fundamental theorem of calculus is used to apply the boundary values to the integral function to find the numerical result. Put the upper limit in the function after that put the lower limit and place the negative sign among them.
The expression for this kind of integral calculus is given below.
mns(z) dz = S(n) – S(m) = M
Let us describe the definite integral with an example in which the rules of integral calculus are involved. The below example will help you to understand how to calculate the numerical value of the function.
Example
Evaluate the definite integral of the given function with respect to z & the [2, 3] is the interval.
s(z) = 1.5z3 – 15.2z5 – 10z4 + 5cos(z) + 150
Solution
Step I:Take the integral function and apply the notation of integration to it.
s(z) = 1.5z3 – 15.2z5 – 10z4 + 5cos(z) + 150
23[s(z)] dz = 23[1.5z3 – 15.2z5 – 10z4 + 5cos(z) + 150] dz
Step II:Use the sum and difference laws of integral calculus and apply the integral notation to each function separately.
23[1.5z3 – 15.2z5 – 10z4 + 5cos(z) + 150] dz = 23[1.5z3] dz – 23[15.2z5] dz – 23[10z4] dz + 23[5cos(z)] dz + 23[150] dz
Step III:Now use the constant function law of integral and take the constant coefficients outside the integral notation.
= 1.523[z3] dz – 15.223[z5] dz – 1023[z4] dz + 523[cos(z)] dz + 15023dz
Step IV: Use the power and trigonometry law of integral calculus and solve the above expression.
= 1.5 [z3+1 / 3 + 1]32 – 15.2 [z5+1 / 5 + 1]32 – 10 [z4+1 / 4 + 1]32 + 5 [sin(z) ]32 + 150 [z]32
= 1.5 [z4 / 4]32 – 15.2 [z6 / 6]32 – 10 [z5 / 5]32 + 5 [sin(z) ]32 + 150 [z]32
= 1.5/4 [z4]32 – 15.2/6 [z6]32 – 10/5 [z5]32 + 5 [sin(z) ]32 + 150 [z]32
= 0.375 [z4]32 – 2.534 [z6]32 – 2 [z5]32 + 5 [sin(z) ]32 + 150 [z]32
Step V:Apply the boundary values.
= 0.375 [34 – 24] – 2.534 [36 – 26] – 2 [35 – 25] + 5 [sin(3) – sin(2)] + 150 [3 – 2]
= 0.375 [81 – 16] – 2.534 [729 – 64] – 2 [243 – 32] + 5 [sin(3) – sin(2)] + 150 [3 – 2]
= 0.375 [81 – 16] – 2.534 [729 – 64] – 2 [243 – 32] + 5 [0.1411 – 0.9093] + 150 [3 – 2]
= 0.375 [65] – 2.534 [665] – 2 [211] + 5 [-0.7682] + 150 [1]
= 24.375 – 1685.11 – 422 + (-3.841) + 150
= -1660.735 – 422 + (-3.841) + 150
= -2082.735 - 3.841 + 150
= -2086.576 + 150
= -1936.576
The other kind of integration that is helpful for finding the new function is known as the indefinite integral. It is frequently used in integral calculus and without using the upper and lower limit values of the function.
You have to integrate the function with respect to the independent variable. The constant of integration is involved in this kind of integral calculus. The expression for this kind of integral calculus is given below.
ʃ s(u) du = S(u) + C
Let us describe the indefinite integral with an example in which the rules of integral calculus are involved. The below example will help you to understand how to calculate the new function.
Example
Evaluate the indefinite integral of with respect to u.
s(u) = 150.3u5 – 20u + 62u3 – sin(u) + 201
Solution
Step I:Take the given integrand and apply the integral notation to it.
s(u) = 150.3u5 – 20u + 62u3 – sin(u) + 201
ʃ s(u) du = ʃ [150.3u5 – 20u + 62u3 – sin(u) + 201] du
Step II:Use the sum and difference laws of integral calculus and apply the integral notation to each function separately.
ʃ [150.3u5 – 20u + 62u3 – sin(u) + 201] du = ʃ [150.3u5] du– ʃ [20u] du + ʃ [62u3] du – ʃ [sin(u)] du + ʃ [201] du
Step III:Now use the constant function law of integral and take the constant coefficients outside the integral notation.
= 150.3ʃ [u5] du– 20ʃ [u] du + 62ʃ [u3] du – ʃ [sin(u)] du + 201ʃ du
Step IV: Use the power and trigonometry law of integral calculus and solve the above expression.
= 150.3 [u5+1 / 5 + 1]– 20 [u1+1 / 1 + 1] + 62 [u3+1 / 3 + 1] – [-cos(u)] + 201 [u] + C
= 150.3 [u6 / 6]– 20 [u2 / 2] + 62 [u4 / 4] – [-cos(u)] + 201 [u] + C
= 150.3/6 [u6]– 20/2 [u2] + 62/4 [u4] – [-cos(u)] + 201 [u] + C
= 150.3/6 [u6]– 10 [u2] + 31/2 [u4] – [-cos(u)] + 201 [u] + C
= 25.05 [u6]– 10 [u2] + 15.5 [u4] – [-cos(u)] + 201 [u] + C
= 25.05u6 – 10u2 + 15.5u4 + cos(u) + 201u + C
Now you can grab all the basics of integral calculus from this post. Now you can solve any integral problem easily by using the rules and formulas of integral calculus.
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