6.5
13
The problem requires finding the area bounded by the curve \( y = |x - 2| \), the x-axis (y = 0), x = 0, and x = 5.
First, analyze the curve:
\(y = |x - 2|\) |
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1. When \(x < 2\), \(y = 2 - x\) |
2. When \(x \geq 2\), \(y = x - 2\) |
The critical point is at \(x = 2\). The expression \(y = |x - 2|\) forms a V-shaped graph.
Calculate the area between x = 0 and x = 5 by dividing it into two regions using x = 2.
For \(0 \leq x < 2\): |
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Area under \(y = 2 - x\) |
\[ \int_{0}^{2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{0}^{2} = \left(4 - 2\right) = 2 \] |
For \(2 \leq x \leq 5\): |
Area under \(y = x - 2\) |
\[ \int_{2}^{5} (x - 2) \, dx = \left[ \frac{x^2}{2} - 2x \right]_{2}^{5} = \left(\frac{25}{2} - 10\right) - \left(\frac{4}{2} - 4\right) = \frac{9}{2} \] |
Summing both areas gives:
Final Area = \( 2 + \frac{9}{2} = \frac{4}{2} + \frac{9}{2} = \frac{13}{2} = 6.5 \) square units.
Hence, the area bounded by the curve is 6.5 square units.
The curve \(y=|x-2|\) is split into two linear parts:
We calculate the area under the curve from \(x = 0\) to \(x = 5\), divided into two regions:
Region 1: \(x \in [0, 2]\) The area under \(y = 2 - x\) is:
\(Area_1 = \int_0^2 (2 - x) dx.\)
\(Area_1 = \left[ 2x - \frac{x^2}{2} \right]_0^2 = \left( 2(2) - \frac{2^2}{2} \right) - \left( 2(0) - \frac{0^2}{2} \right).\)
\(Area_1 = (4 - 2) - 0 = 2.\)
Region 2: \(x \in [2, 5]\) The area under \(y = x - 2\) is:
\(Area_2 = \int_2^5 (x - 2) dx.\)
\(Area_2 = \left[ \frac{x^2}{2} - 2x \right]_2^5 = \left( \frac{5^2}{2} - 2(5) \right) - \left( \frac{2^2}{2} - 2(2) \right).\)
\(Area_2 = \left( \frac{25}{2} - 10 \right) - \left( \frac{4}{2} - 4 \right).\)
\(Area_2 = \left( \frac{25}{2} - \frac{20}{2} \right) - \left( \frac{4}{2} - \frac{8}{2} \right) = \frac{5}{2} + \frac{4}{2} = \frac{9}{2}.\)
Total Area:
\(Total Area = Area_1 + Area_2 = 2 + \frac{9}{2} = \frac{4}{2} + \frac{9}{2} = \frac{13}{2} = 6.5.\)
Thus, the area bounded by the curve is 6.5 square units.