**Introduction to Area Under a Curve**

The primary application of definite integration is to determine the area under a curve. This area is bounded by the curve, the x-axis, and two vertical lines corresponding to the limits of integration.

**Positive Area**: When the function lies above the x-axis between the two points, the area is positive. This is because the function's values are positive, and thus, the curve is above the x-axis.**Negative Area**: Conversely, if the function lies below the x-axis, the area is taken as negative. This is indicative of the function having negative values in that interval.

**Example:**

Consider the function f(x) = x^{2}. To evaluate the area under this curve from x = 0 to x = 2, we integrate the function over this interval.

The result of the integration is x^{3}/3 evaluated from 0 to 2. This gives a value of 8/3, representing the area under the curve between these two points.

Understanding the area under a curve is crucial for grasping the concept of radians in trigonometry.

**Properties of Definite Integration**

Definite integration is governed by several properties that simplify its application and provide deeper insights into the nature of functions. Here are some pivotal properties:

1. **Additivity over Intervals**: If you're integrating a function over an interval [a, c], and there's a point b such that a < b < c, then the integral from a to c is the sum of the integrals from a to b and from b to c. This property allows us to break down complex regions into simpler shapes that are easier to integrate.

2.** Reversing Limits**: An interesting property is that if you swap the limits of integration, the result is the negative of the original integral. This is because you're effectively reversing the direction in which you're accumulating area.

3. I**ntegrating Zero**: If you integrate the function zero over any interval, the result is always zero. This is intuitive since the graph of the function zero is the x-axis itself, and there's no area to accumulate.

4.** Multiplicative Scalar**: If you multiply a function by a scalar (a constant) and then integrate it, it's the same as integrating the function first and then multiplying the result by the scalar.

5. **Additivity over Functions**: If you have two functions, the integral of their sum is the sum of their integrals. This property is a direct result of how areas add up geometrically.

Understanding these properties can also aid in finding local extrema in functions, which is another essential calculus skill.

**Example:**

To evaluate the integral of the function 2x + 3 from x = 2 to x = 4, we can split the function into its constituent parts and integrate each one. The result is a value of 18, representing the combined area under the curve of both functions in the given interval.

For more complex functions, knowing the product and quotient rules can be very beneficial.

**Deep Dive into Area Under a Curve**

The concept of area under a curve is foundational in various scientific fields:

**Physics**: In physics, the area under a velocity-time graph gives the displacement. Similarly, the area under a force-distance graph provides the work done.**Economics**: In economics, the area under a marginal cost curve up to a certain quantity gives the total variable cost up to that quantity.**Medicine**: In pharmacology, the area under a concentration-time graph gives the total drug exposure over time.

**Example:**

In a business context, if a company's marginal cost function (cost to produce one additional unit) is given by C'(x) = 3x^{2} + 4x, the total cost to produce 3 units, excluding fixed costs, can be found by integrating this function from 0 to 3. The result is £45, which, when added to any fixed costs, gives the total cost of production.

For further understanding, reviewing the equation of a tangent line can provide insights into how slopes of curves are calculated.

Additionally, exploring the applications of Bayes' theorem can offer a deeper understanding of probability in conjunction with calculus.

## FAQ

Definite and indefinite integrations are both forms of integration, but they serve different purposes. Indefinite integration, represented by ∫f(x) dx, yields a family of functions (plus a constant of integration) known as antiderivatives. It does not have specified limits of integration. On the other hand, definite integration, represented by ∫ from a to b of f(x) dx, provides a specific numerical value. It represents the net area under the curve f(x) from x = a to x = b. The result can be positive, negative, or zero based on the function and the interval.

Properties of definite integrals, such as additivity over intervals, reversing limits, and additivity over functions, offer shortcuts and techniques to simplify the process of integration. For instance, if a function is symmetric about the y-axis, its integral from -a to a can be easily determined by doubling its integral from 0 to a. Similarly, if you know the integral value over one interval, the additivity property can help deduce it over another interval without performing the integration again. Utilising these properties can save time and reduce the complexity of problems.

The Fundamental Theorem of Calculus establishes a connection between differentiation and integration. It states that if a function is continuous over an interval and has an antiderivative F, then the definite integral of the function over that interval can be computed using the antiderivative's values at the endpoints. Specifically, the integral from a to b of f(x) dx is equal to F(b) - F(a). This theorem provides a practical way to evaluate definite integrals without having to compute the area under the curve directly.

A negative value for a definite integral indicates that the majority of the curve lies below the x-axis in the interval of integration. In geometric terms, it means that the area being considered is primarily between the curve and the x-axis but below the x-axis. It's essential to understand that the integral calculates the "net area". If the curve dips below the x-axis, the area below is taken as negative. In real-world applications, this could represent losses, deficits, or any other scenario where a decrease or negative value is implied.

Yes, definite integrals can be employed to determine volumes. When you want to find the volume of a solid of revolution (a shape obtained by rotating a curve around an axis), you can use definite integration. By taking infinitesimally thin discs or shells, finding their volume, and then summing these volumes over an interval, you get the total volume of the solid. The method used, whether discs, washers, or cylindrical shells, depends on the specifics of the problem and the axis of rotation.

## Practice Questions

To evaluate the definite integral of the function f(x) = 3x^{2} + 2x from x = 1 to x = 4, we integrate the function over this interval. The antiderivative of f(x) is x^{3} + x^{2}. Evaluating this antiderivative at the upper limit and subtracting its value at the lower limit, we get: Integral from 1 to 4 of (3x^{2} + 2x) dx = [x^{3} + x^{2}] from 1 to 4 = (64 + 16) - (1 + 1) = 78 Thus, the value of the definite integral is 78.

To find the area bounded by the curve y = 4x - x^{2}, the x-axis, and the lines x = 0 and x = 4, we need to evaluate the definite integral of the function from 0 to 4. The antiderivative of y is 2x^{2} - (1/3)x^{3}. Evaluating this antiderivative at the upper limit and subtracting its value at the lower limit, we get: Integral from 0 to 4 of (4x - x^{2}) dx = [2x^{2} - (1/3)x^{3}] from 0 to 4 = (32 - 64/3) - 0 = 32/3 Thus, the area under the curve is 32/3 square units.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.