https://file.aiquickdraw.com/mathbot/file/17267723292884teudo7n.jpg

Let's analyze the graph step by step in terms of areas to evaluate the integrals.

### Understanding the Graph
The graph consists of straight line segments, so we can break it into geometric shapes (triangles and rectangles) and calculate the areas under or above the x-axis.

#### (a) \(\int_0^{18} f(x) \, dx\)
From \(x = 0\) to \(x = 18\), the graph forms a right triangle with a base of 18 units and a height of 18 units. The area of this triangle is:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 18 \times 18 = 162
\]
The integral \(\int_0^{18} f(x) \, dx = 162\), and the solution provided earlier (324) seems to be incorrect here.

#### (b) \(\int_0^{45} f(x) \, dx\)
We need to sum the areas from \(x = 0\) to \(x = 45\):
1. Triangle from \(x = 0\) to \(x = 18\): Area = 162 (as calculated).
2. Rectangle from \(x = 18\) to \(x = 36\): Area = \(18 \times 18 = 324\).
3. Triangle from \(x = 36\) to \(x = 45\) (below the x-axis): Base = 9, Height = 18, Area = \(\frac{1}{2} \times 9 \times 18 = 81\), but since it is below the x-axis, the area contributes negatively: \(-81\).

Summing the areas:
\[
162 + 324 - 81 = 405
\]

#### (c) \(\int_{45}^{63} f(x) \, dx\)
From \(x = 45\) to \(x = 63\):
- Triangle from \(x = 45\) to \(x = 54\): Base = 9, Height = 18, Area = \(-81\) (negative because it's below the x-axis).
- Triangle from \(x = 54\) to \(x = 63\): Base = 9, Height = 18, Area = \(\frac{1}{2} \times 9 \times 18 = 81\) (positive as it’s above the x-axis).

Summing these areas:
\[
-81 + (-81) = -162
\]
The previously entered answer of -243 does not match the computed value, which should be \(-162\).

#### (d) \(\int_0^{81} f(x) \, dx\)
For the interval \(x = 0\) to \(x = 81\), we sum the previous calculations:
- \(\int_0^{18} f(x) \, dx = 162\)
- \(\int_{18}^{45} f(x) \, dx = 243\)
- \(\int_{45}^{63} f(x) \, dx = -162\)
- Triangle from \(x = 63\) to \(x = 81\) (below x-axis): Base = 18, Height = 18, Area = \(-\frac{1}{2} \times 18 \times 18 = -162\).

Total:
\[
162 + 243 - 162 - 162 = 81
\]

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Would you like further details on any of these steps or areas? 

Here are some related questions:
1. How are integrals related to geometric areas under curves?
2. How do positive and negative areas affect definite integrals?
3. How can we decompose more complex graphs into simple geometric shapes?
4. How can symmetries in a function simplify integral calculations?
5. How does the concept of net area differ from total area?

**Tip:** Always pay attention to whether a region is above or below the x-axis when interpreting integrals as areas.

The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.

Learn how to evaluate definite integrals by interpreting them as areas under or above a graph. This example uses geometric shapes like triangles and rectangles to compute the integral values step by step.

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