**Introduction to Integration**

Integration can be thought of as the process of calculating the area under a curve, formed by a function f(x), and the x-axis. The integral of a function can be represented symbolically as:

Integral f(x) dx

Where:

- Integral symbolises the integral.
- f(x) is the function to be integrated.
- dx indicates the variable of integration.

**Power Rule**

The power rule for integration states that:

Integral x^{n} dx = (x^{(n + 1)}) / (n + 1) + C

Where:

- n is a real number.
- C is the constant of integration.

#### Example 1:

Evaluate Integral x^{3} dx.

Using the power rule:

Integral x^{3} dx = (x^{(3 + 1)}) / (3 + 1) + C = (x^{4}) / 4 + C

**Constant Rule**

The constant rule for integration states that:

Integral k dx = kx + C

Where:

- k is a constant.
- C is the constant of integration.

#### Example 2:

Evaluate Integral 5 dx.

Using the constant rule:

Integral 5 dx = 5x + C

**Applications of Basic Integration Rules**

Understanding the basic rules of integration provides a foundation for solving problems related to areas under curves and other applications in physics, economics, and various scientific fields.

**Finding the Area Under a Curve**

The integral can be used to find the area under the curve of a function between two points (a and b) on the x-axis. The formula to find this area is:

A = Integral from a to b f(x) dx

Where:

- A is the area.
- a and b are the limits of integration.

#### Example 3:

Find the area under the curve y = x^{2} from x = 0 to x = 2.

A = Integral from 0 to 2 x^{2} dx

Using the power rule:

= (x^{(2 + 1)}) / (2 + 1) | from 0 to 2 = (x^{3}) / 3 | from 0 to 2 = (2^{3}) / 3 - (0^{3}) / 3 = 8 / 3

**Solving Physical Problems**

Integration is pivotal in physics, especially in kinematics, to find quantities like displacement and velocity when given their respective rates.

#### Example 4:

If a particle moves along a line with velocity v(t) = t^{2}, find the displacement between t = 1 and t = 3.

S = Integral from 1 to 3 t^{2} dt

Using the power rule:

= (t^{(2 + 1)}) / (2 + 1) | from 1 to 3 = (t^{3}) / 3 | from 1 to 3 = (3^{3}) / 3 - (1^{3}) / 3 = 8

Thus, the particle is displaced by 8 units from t = 1 to t = 3.

**Practice Questions**

**Question 1:**

Evaluate Integral 4x^{5} dx.

**Question 2:**

Find the area under the curve y = 3x from x = 0 to x = 4.

**Question 3:**

If a particle has an acceleration of a(t) = 6t, find the velocity function, given that the velocity is 5 m/s when t = 0.

**Question 4:**

Evaluate Integral (2x^{3} - 3x^{2} + 4) dx.

**Question 5:**

Find the area under the curve y = x^{3} from x = -1 to x = 1.

## FAQ

A negative integral does not imply an error or lack of solution; rather, it has a clear geometric interpretation. When we calculate the integral of a function and obtain a negative result, it indicates that the majority of the area under the curve lies below the x-axis. In other words, if f(x) is negative over the interval [a, b], then the integral from a to b of f(x) will be negative. This is because the y-values (function values) are negative, and when summed, yield a negative total area. It's crucial to understand that this negative sign indicates direction or orientation, not a lack of area.

No, not all functions can be integrated using basic integration rules like the power rule and constant rule. Some functions require more advanced techniques of integration, such as substitution, partial fractions, and integration by parts, to find their antiderivatives. Moreover, there are functions that do not have an elementary antiderivative, meaning they cannot be expressed in terms of basic functions. In such cases, numerical methods or special functions, like the error function (erf), might be used to approximate or represent the integral.

The constant rule of integration, Integral k dx = kx + C, where k is a constant, is widely applicable in real-world scenarios, particularly in physics and engineering. For instance, when calculating the work done by a constant force over a certain distance, the constant rule is applied since work is the integral of force with respect to distance. Similarly, in economics, when calculating total cost given a constant rate of cost, the constant rule is used to integrate the cost rate over a certain quantity. This rule simplifies problems where quantities change at a constant rate, providing straightforward solutions in various fields.

Integration is fundamentally connected to finding the area under a curve because the integral essentially accumulates quantities. When we integrate a function f(x) from a to b, we are summing up all the tiny areas under the curve f(x) from x = a to x = b. The integral calculates the net area, considering areas above the x-axis as positive and areas below as negative. This concept is widely used in physics and engineering to calculate quantities like work done, where the force (curve) is integrated over a distance, effectively summing up all the infinitesimally small amounts of work done along the path.

When we find the integral of a function, we are essentially finding the antiderivative, meaning a function whose derivative gives the original function. The constant of integration, often denoted as C, is added to account for the fact that the derivative of a constant is zero. Therefore, when we differentiate our antiderivative, the constant disappears and we are left with our original function. Including C ensures that we account for all possible antiderivatives, as any constant could have been present in the original function before differentiation. It's crucial in accurately representing all possible solutions to an integral.

## Practice Questions

To evaluate the integral of the function f(x) = 3x^{2} + 2x + 1 from x = 1 to x = 4, we will apply the power rule and constant rule of integration. First, we find the antiderivative (indefinite integral) of f(x):

Integral f(x) dx = Integral (3x^{2} + 2x + 1) dx = (x^{3}) + (x^{2}) + x + C

Now, we evaluate the definite integral from 1 to 4:

F(4) - F(1) = ((4^{3}) + (4^{2}) + 4) - ((1^{3}) + (1^{2}) + 1) = (64 + 16 + 4) - (1 + 1 + 1) = 84 - 3 = 81

Therefore, the integral of f(x) = 3x^2 + 2x + 1 from x = 1 to x = 4 is 81.

To find the total displacement of the particle from t = 1 to t = 3, we need to evaluate the integral of the velocity function v(t) = t^{2} + 2t from 1 to 3. The integral of the velocity function will give us the displacement function, which we can use to find the total displacement in the given time interval.

Integral from 1 to 3 v(t) dt = Integral from 1 to 3 (t^{2} + 2t) dt = [ (t^{3})/3 + t^{2} ] from 1 to 3 = [ (3^{3})/3 + (3^{2}) ] - [ (1^{3})/3 + (1^{2}) ] = [ 9 + 9 ] - [ 1/3 + 1 ] = 18 - 4/3 = 54/3 - 4/3 = 50/3

Thus, the total displacement of the particle from t = 1 to t = 3 is 50/3 units.