# How to solve problems of integral calculus? Explained with examples

Calculus is a general branch of mathematics and is used in various branches to do different engineering and daily life tasks. The calculus is further divided into various sub-types such as differentiation, integration, limit, etc.

The process of finding complex calculus problems is rather tough as various kinds of algebraic and trigonometric functions are used. The process of differentiation is easy than integration. In this post, we will learn how to solve advanced integral problems along with examples.

## What is integral?

In calculus, the integral is simply defined as the process of finding the integral of the function or first derivative. The term integral is widely used to evaluate the area and volume.

The reverse process of derivative to calculate the original function is said to be the antiderivative in calculus. The integral can also be defined as finding the new function whose original function is derivative or calculating the numerical values by applying the upper and lower limit values.

## Types of integral in calculus

There are two ways of calculating the integral of the function. First is to find a function whose original function is the first derivative while the other way is to fund the area or numerical values of the function through a graph. On the basis of these conditions, the integral is divided into types.

1. Definite integral
2. Indefinite integral

### Definite integral

The first type of integral is definite integral in which you have to find the definite result of the given function under the given conditions such as using the upper and lower limit values with the help of a fundamental theorem of calculus.

According to the fundamental theorem of calculus, first, find the new function and then apply the upper limit value minus the lower limit value.

ba f(x) dx = F(b) – F(a)

### Indefinite integral

The other type of integral is indefinite integral in which you have to evaluate the new function from the differential function. In this type of integral, the boundary values are not used. The result of integral in this type is indefinite along with an arbitrary integrating constant C.

∫ f(x) dx = F(x)

## Formulas of Integral calculus

Below are some formulas and rules of integral calculus used to evaluate advanced integral problems.

### Power rule

∫ yn dy = yn+1 /n+1 + C, where n ≠ -1

∫ dy = y + C

### Trigonometric formulas

1. ∫ cos(y) dy = sin(y) + C
2. ∫ sin(y) dy = -cos(y) + C
3. ∫ sec2x dx = tanx + C
4. ∫ cosec2(y) dy = -cot(y) + C
5. ∫ sec2(y) dy = tan(y) + C
6. ∫ sec(y) * tan(y) dy = sec(y) + C
7. ∫ csc(y) * cot(y) dy = -csc(y) + C

An integration calculator By Meracalculator can be a helpful source to evaluate the problems of integral calculus according to the above rules and formulas.

## How to evaluate the problems of integral calculus?

The advance and complex integral problems can be evaluated easily with the help of formulas and rules of integral calculus. The advanced problems involved integral and double integral problems.

### Example 1: For integral

Evaluate the first integral of the given function with respect to “u”.

f(u) = 12u3 + 6sin(u) + 4sec2(u) – 5u4 + 20

Solution

Step 1: Take the given function and write it according to the general expression of the integral.

f(u) = 12u3 + 6sin(u) + 4sec2(u) – 5u4 + 20

ʃ f(u) du = ʃ [12u3 + 6sin(u) + 4sec2(u) – 5u4 + 20] du

Step 2: Apply the sum and difference laws of integration and write the notation of integral separately with each term of the above expression.

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = ʃ [12u3] du + ʃ [6sin(u)] du + ʃ [4sec2(u)] du – ʃ [5u4] du + ʃ  du

Step 3: Now apply the constant function rule to the above expression.

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 12ʃ [u3] du + 6ʃ [sin(w)] du + 4ʃ [sec2(u)] du – 5ʃ [u4] du + ʃ  du

Step 4: Now use the trigonometric and power laws of integration to integrate the above expression.

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 12 [u3+1 / 3 + 1] + 6 [-cos(u)] + 4 [tan(u)] – 5 [u4+1 / 4 + 1] + [20u] + C

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 12 [u4 / 4] + 6 [-cos(u)] + 4 [tan(u)] – 5 [u5 / 5] + [20u] + C

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 12/4 [u4] + 6 [-cos(u)] + 4 [tan(u)] – 5/5 [u5] + [20u] + C

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 3 [u4] + 6 [-cos(u)] + 4 [tan(u)] – [u5] + [20u] + C

ʃ [12u3 + 6sin(w) + 4sec2(u) – 5u4 + 20] du = 3u4 – 6cos(u) + 4tan(u) – u5 + 20u + C

### Example 2: For double integral

Evaluate the double integral of the given function with respect to “w & z”.

f(w, z) = 4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2

Solution

Step 1: Take the given function and write it according to the general expression of the integral.

f(w, z) = 4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2

ʃ ʃ f(w, z) dw dz = ʃ ʃ [4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw dz

Step 2: Now integrate the above expression with respect to “w”.

ʃ ʃ [4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw dz = ʃ{ ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw} dz … (1)

for w

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = ʃ [4w3z] dw + ʃ [4w5] dw – ʃ [4z4] dw – ʃ [6w2z2] dw + ʃ [4wz2] dw

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 4zʃ [w3] dw + 4ʃ [w5] dw – 4z4ʃ [] dw – 6z2ʃ [w2] dw + 4z2ʃ [w] dw

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 4z [w3+1 / 3 + 1] + 4 [w5+1 / 5 + 1] – 4z4 [w] – 6z2 [w2+1 / 2 + 1] + 4z2 [w1+1 / 1 + 1]

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 4z [w4 / 4] + 4 [w6 / 6] – 4z4 [w] – 6z2 [w3 / 3] + 4z2 [w2 / 2]

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 4z/4 [w4] + 4/6 [w6] – 4z4 [w] – 6z2/3 [w3] + 4z2/2 [w2]

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 2z [w4] + 2/3 [w6] – 4z4 [w] – 2z2 [w3] + 2z2 [w2]

ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw = 2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2

Step 3: Put the above result in (1) and then evaluate the expression with respect to z.

ʃ ʃ [4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw dz = ʃ{ ʃ[4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw} dz … (1)

ʃ ʃ [4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw dz = ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz

for z

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = ʃ [2zw4] dz + ʃ [2w6/3] dz – ʃ [4wz4] dz – ʃ [2z2w3] dz + ʃ [2z2w2] dz

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = 2w4 ʃ [z] dz + 2w6/3ʃ [] dz – 4w ʃ [z4] dz – 2w3 ʃ [z2] dz + 2 w2ʃ [z2] dz

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = 2w4 [z1+1 / 1 + 1] + 2w6/3 [z] – 4w [z4+1 / 4 + 1] – 2w3 [z2+1 / 2 + 1] + 2 w2 [z2+1 / 2 + 1] + C

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = 2w4 [z2 / 2] + 2w6/3 [z] – 4w [z5 / 5] – 2w3 [z3 / 3] + 2 w2 [z3 / 3] + C

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = 2w4z2/2 + 2w6z/3 – 4wz5/5 – 2w3z3/3 + 2w2z3/3 + C

ʃ [2zw4 + 2w6/3 – 4wz4 – 2z2w3 + 2z2w2] dz = w4z2 + 2w6z/3 – 4wz5/5 – 2w3z3/3 + 2w2z3/3 + C

Step 4: Final result is:

ʃ ʃ [4w3z + 4w5 – 4z4 – 6w2z2 + 4wz2] dw dz = w4z2 + 2w6z/3 – 4wz5/5 – 2w3z3/3 + 2w2z3/3 + C

## Conclusion

In this blog post, we have covered all the basics of integration and how to solve advanced integral problems with examples. Now you can grab all the basics of solving integral and double integral problems by following the above-mentioned procedure.