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Chapter 6: Problem 9

Find the area of the region between the curve and the \(x\) -axis. $$f(x)=x^{2}-2 x-3, \text { from } 0 \text { to } 2$$

Short Answer

Expert verified
The area is \(-\frac{22}{3}\).

Step by step solution

01

- Set Up the Definite Integral

To find the area between the curve and the x-axis from 0 to 2, we use a definite integral: \[\text{Area} = \int_{0}^{2} (f(x)) \, dx \] in this case, \[f(x) = x^2 - 2x - 3 \]
02

- Integrate the Function

Calculate the integral of \(f(x)\): \[\text{Integral} = \int_{0}^{2} (x^2 - 2x - 3) \, dx \] The antiderivative of \(x^2 - 2x - 3\) is \(\frac{x^3}{3} - x^2 - 3x\)
03

- Evaluate the Integral at the Bounds

Apply the bounds 0 to 2 to the antiderivative result: \[\left. \left( \frac{x^3}{3} - x^2 - 3x \right) \right|_{0}^{2} \] Substitute the bounds into the antiderivative: \[\left( \frac{2^3}{3} - 2^2 - 3(2) \right) - \left( \frac{0^3}{3} - 0^2 - 3(0) \right) \]
04

- Simplify the Expression

Simplify the expression: \[\left( \frac{8}{3} - 4 - 6 \right) - \left( 0 \right) = \frac{8}{3} - 10 \] Calculate the value: \[\frac{8}{3} - 10 = \frac{8 - 30}{3} = -\frac{22}{3} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
When we talk about the area between a curve and the x-axis, we are referring to a 'definite integral'. A definite integral is a type of integral that has specific limits or bounds, in this case from 0 to 2. It helps us calculate the total area under a curve between those two points.
The definite integral can be written as: \(\int_{a}^{b} f(x) \, dx\).
Here, \(a\) and \(b\) are the bounds, and \(f(x) \) is the function.
So for our problem: \(\int_{0}^{2} (x^2 - 2x - 3) \, dx\).
Antiderivative
In order to solve the definite integral, we need to find the antiderivative of the function, \(f(x) = x^2 - 2x - 3\).
The antiderivative, or the indefinite integral, is the opposite of differentiation.
For \(x^2\), the antiderivative is \(\frac{x^3}{3}\).
For \(-2x\), the antiderivative is \(-x^2\).
For \(-3\), the antiderivative is \(-3x \).
Combining these, the antiderivative of \(x^2 - 2x - 3 \) is \(\frac{x^3}{3} - x^2 - 3x \).
Bounds Evaluation
After finding the antiderivative, our next step is to evaluate it at the given bounds. These are the limits of integration, which are 0 and 2 in this case.
We substitute each bound into the antiderivative and take the difference.
Mathematically, this is written as: \(\big[ \frac{x^3}{3} - x^2 - 3x \big]_{0}^{2}\).
So, it becomes \(\big( \frac{2^3}{3} - 2^2 - 3(2) \big) - \big( \frac{0^3}{3} - 0^2 - 3(0) \big)\).
Simplifying it: \( \big( \frac{8}{3} - 4 - 6 \big) - (0) = \frac{8}{3} - 10 \).
Function Integration
Putting all the steps together, we use the integral of the function to evaluate the area.
First, set up the definite integral: \(\text{Area} = \int_{0}^{2} (x^2 - 2x - 3) \, dx\).
Next, find the antiderivative of the function: \(\frac{x^3}{3} - x^2 - 3x\).
Then, evaluate the antiderivative at the bounds: \(\big( \frac{8}{3} - 4 - 6 \big) - 0 = \frac{8}{3} - 10 \).
Finally, simplify the expression: \(\frac{8}{3} - 10 = \frac{8 - 30}{3} = -\frac{22}{3} \).
Thus, the area between the curve and the x-axis from 0 to 2 ends up being \(-\frac{22}{3} \). Notice the negative value indicates the curve is below the x-axis between these bounds.

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