Sketch the graph y  x  2 . Evaluate  x  2dx . What does the value of the integral represent on the

0

graph?

Let's tackle this problem step by step.

### 1. **Sketch the Graph of \( y = x - 2 \)**

The function \( y = x - 2 \) is a linear equation with a slope of 1 and a y-intercept at \(-2\).

- The slope \( m = 1 \) means that for every unit increase in \( x \), \( y \) increases by 1.
- The y-intercept is \(-2\), so the line crosses the y-axis at the point \( (0, -2) \).

To sketch the graph:
- Start at the point \( (0, -2) \).
- Since the slope is 1, from the point \( (0, -2) \), go up 1 unit and right 1 unit to get the next point \( (1, -1) \).
- Continue this process to draw the line.

### 2. **Evaluate the Integral \( \int_0^x (x - 2) \, dx \)**

The integral is given by:

\[
\int (x - 2) \, dx
\]

This can be integrated term by term:

\[
\int x \, dx - \int 2 \, dx
\]

- The integral of \( x \) with respect to \( x \) is:

\[
\frac{x^2}{2}
\]

- The integral of 2 with respect to \( x \) is:

\[
2x
\]

So the indefinite integral is:

\[
\int (x - 2) \, dx = \frac{x^2}{2} - 2x + C
\]

Where \( C \) is the constant of integration.

Now, to evaluate this integral over a specific interval, say from \( 0 \) to \( a \), we compute:

\[
\int_0^a (x - 2) \, dx = \left[\frac{x^2}{2} - 2x\right]_0^a
\]

Substitute the limits:

\[
\left(\frac{a^2}{2} - 2a\right) - \left(\frac{0^2}{2} - 2(0)\right) = \frac{a^2}{2} - 2a
\]

So the evaluated definite integral from \( 0 \) to \( a \) is:

\[
\int_0^a (x - 2) \, dx = \frac{a^2}{2} - 2a
\]

### 3. **Interpretation of the Integral on the Graph**

The definite integral \( \int_0^a (x - 2) \, dx \) represents the area under the curve \( y = x - 2 \) from \( x = 0 \) to \( x = a \) and above the x-axis.

- If the curve is above the x-axis, the integral gives the area under the curve.
- If the curve is below the x-axis, the integral gives the negative of the area above the curve to the x-axis.

Since the line \( y = x - 2 \) crosses the x-axis at \( x = 2 \), the integral from \( 0 \) to \( a \) will depend on the value of \( a \):

- For \( 0 \leq a < 2 \), the line is below the x-axis, and the integral will be negative, indicating the area below the x-axis.
- For \( a > 2 \), the integral has a positive and a negative part.

### Example for \( a = 2 \):

If \( a = 2 \), the integral \( \int_0^2 (x - 2) \, dx \) is:

\[
\frac{2^2}{2} - 2(2) = \frac{4}{2} - 4 = 2 - 4 = -2
\]

This negative value represents the area of the triangle formed below the x-axis between \( x = 0 \) and \( x = 2 \).

Would you like to explore this further or have any other questions?

### Related Questions:
1. How do you find the area between two curves?
2. What happens to the integral if the limits are swapped?
3. How does the interpretation of the integral change if the curve is nonlinear?
4. What are the applications of definite integrals in real life?
5. How does the integral of a function relate to the function's antiderivative?

### Tip:
Remember, the definite integral represents the net area between the curve and the x-axis, considering the sign (positive for above and negative for below).

Learn how to sketch the graph of y = x - 2 and evaluate the integral ∫(x - 2) dx from x = 0 to x = a. Understand the interpretation of the integral as the area under the curve y = x - 2 and above the x-axis. Suitable for advanced high school students.

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