We are tasked with evaluating the integral:

$\int_0^9 |2x - 9| \, dx$

This integral involves the absolute value function, which means we need to handle the function differently based on its behavior.

Step 1: Understanding the Absolute Value Function

The expression inside the absolute value is $2x - 9$. The absolute value function behaves as follows:

\begin{cases} 2x - 9 & \text{if } 2x - 9 \geq 0 \quad \text{(i.e., for } x \geq 4.5 \text{)} \\ -(2x - 9) & \text{if } 2x - 9 < 0 \quad \text{(i.e., for } x < 4.5 \text{)} \end{cases}$$ This means that we need to split the integral at $$x = 4.5$$ (since $$2x - 9 = 0$$ at $$x = 4.5$$). ### Step 2: Splitting the Integral We now split the integral into two parts, one from $$0$$ to $$4.5$$ and the other from $$4.5$$ to $$9$$: $$\int_0^9 |2x - 9| \, dx = \int_0^{4.5} -(2x - 9) \, dx + \int_{4.5}^9 (2x - 9) \, dx$$ ### Step 3: Evaluating Each Integral #### First Integral: From 0 to 4.5 $$\int_0^{4.5} -(2x - 9) \, dx = \int_0^{4.5} (9 - 2x) \, dx$$ The integral of $$9 - 2x$$ is: $$\int (9 - 2x) \, dx = 9x - x^2$$ Evaluating from $$0$$ to $$4.5$$: $$[9x - x^2]_0^{4.5} = \left( 9(4.5) - (4.5)^2 \right) - \left( 9(0) - (0)^2 \right)$$ $$= 40.5 - 20.25 = 20.25$$ #### Second Integral: From 4.5 to 9 $$\int_{4.5}^9 (2x - 9) \, dx$$ The integral of $$2x - 9$$ is: $$\int (2x - 9) \, dx = x^2 - 9x$$ Evaluating from $$4.5$$ to $$9$$: $$[x^2 - 9x]_{4.5}^9 = \left( 9^2 - 9(9) \right) - \left( (4.5)^2 - 9(4.5) \right)$$ $$= (81 - 81) - (20.25 - 40.5) = 0 + 20.25 = 20.25$$ ### Step 4: Final Answer Adding both integrals: $$\text{Total Area} = 20.25 + 20.25 = 40.5$$ Thus, the value of the integral is: $$\boxed{40.5}$$ --- Would you like further details on any part of this solution? Here are some related questions you might find useful: 1. How would the result change if the limits were from 0 to 5 instead of 0 to 9? 2. Can this method be used to integrate other absolute value functions? How? 3. How does the concept of areas apply to definite integrals involving absolute values? 4. What happens if the integrand has more than one absolute value term? 5. How would you approach a similar problem with a quadratic function inside the absolute value? **Tip:** When dealing with absolute value integrals, always split the integral at points where the expression inside the absolute value equals zero to avoid complications.