91Ó°ÊÓ

A definite integral is like finding the total sum or area under a curve from one point to another. Imagine you are shading the region under the curve from the starting point to the ending point of an interval on a graph.
This specific area can be calculated using the definite integral. In mathematical terms, for a function \( f(x) \) over an interval \([a, b]\), the definite integral is written as:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \( a \) and \( b \) are the limits of integration. They represent the start and end of the interval. In our example, we want to find the area under the curve \( y=(x-0.5)^2+1.75 \) from \( x = -1 \) to \( x = 2 \). So, our definite integral becomes:

• \( \int_{-1}^{2} ((x-0.5)^2 + 1.75) \, dx \)

By working this out, we get the area under the curve over the specified interval.

The power rule for integration is a tool that helps us find integrals easily, especially when dealing with polynomials. The rule states that the integral of \( x^n \), where \( n e -1 \), is:

• \( \frac{x^{n+1}}{n+1} + C \)

Here, \( C \) is the constant of integration, used mainly in indefinite integrals.
In our problem, you use this rule to integrate \((x-0.5)^2\):

• \( \frac{(x-0.5)^3}{3} \)

Since our integral also involves a constant function \(1.75\), we integrate it as \(1.75x\). These integrated forms help us construct the antiderivative needed to evaluate the definite integral.
The resulting expression is:

• \( \left[ \frac{(x-0.5)^3}{3} + 1.75x \right] \)

With this expression, we can substitute the limits.

The concept of the area under a curve refers to the space beneath a curve and above the x-axis within a specific interval. This area is crucial in understanding various real-world scenarios, such as calculating distance, volume, and more.
In the context of calculus, this area is found using definite integrals. For our problem, after setting up the integral using the power rule, the next step was evaluating it by substituting the interval limits into the antiderivative.

• Substitute the upper limit: \( \left[ \frac{(2-0.5)^3}{3} + 1.75(2) \right] \)
• Substitute the lower limit: \( \left[ \frac{(-1-0.5)^3}{3} + 1.75(-1) \right] \)
• Subtract the lower limit result from the upper limit result.

This subtraction gives the total area under the curve from \( x = -1 \) to \( x = 2 \), which is \( 2.125 \).
Visualize this process like peeling off a slice of pie, measuring its area, and knowing exactly how much pie you have!