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Question
Math
Posted 9 months ago

The graph hh is shown below.

Evaluate the definite integral of ∫−9−3h(x)dx\int_{-9}^{-3} h(x) d x.
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Answer from Sia
Posted 9 months ago
Solution by Steps
step 1
To evaluate the definite integral ∫−9−3h(x) dx\int_{-9}^{-3} h(x) \, dx, we need to find the area under the curve of h(x)h(x) from x=−9x = -9 to x=−3x = -3
step 2
From the graph, we observe that h(x)h(x) is a piecewise linear function. We need to break the integral into segments where h(x)h(x) is linear
step 3
The first segment is from x=−9x = -9 to x=−8x = -8. The function h(x)h(x) is linear and can be described by the line connecting the points (−12,0)(-12, 0) and (−8,5)(-8, 5)
step 4
The second segment is from x=−8x = -8 to x=−6x = -6. The function h(x)h(x) is linear and can be described by the line connecting the points (−8,5)(-8, 5) and (−6,0)(-6, 0)
step 5
The third segment is from x=−6x = -6 to x=−3x = -3. The function h(x)h(x) is linear and can be described by the line connecting the points (−6,0)(-6, 0) and (−3,−4)(-3, -4)
step 6
Calculate the area of each segment using the formula for the area of a trapezoid: Area=12×(Base1+Base2)×Height\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
step 7
For the segment from x=−9x = -9 to x=−8x = -8: Base1=h(−9)=0\text{Base}_1 = h(-9) = 0, Base2=h(−8)=5\text{Base}_2 = h(-8) = 5, and Height=1\text{Height} = 1. So, Area1=12×(0+5)×1=2.5\text{Area}_1 = \frac{1}{2} \times (0 + 5) \times 1 = 2.5
step 8
For the segment from x=−8x = -8 to x=−6x = -6: Base1=h(−8)=5\text{Base}_1 = h(-8) = 5, Base2=h(−6)=0\text{Base}_2 = h(-6) = 0, and Height=2\text{Height} = 2. So, Area2=12×(5+0)×2=5\text{Area}_2 = \frac{1}{2} \times (5 + 0) \times 2 = 5
step 9
For the segment from x=−6x = -6 to x=−3x = -3: Base1=h(−6)=0\text{Base}_1 = h(-6) = 0, Base2=h(−3)=−4\text{Base}_2 = h(-3) = -4, and Height=3\text{Height} = 3. So, Area3=12×(0+−4)×3=−6\text{Area}_3 = \frac{1}{2} \times (0 + -4) \times 3 = -6
step 10
Sum the areas of all segments to find the total area under the curve from x=−9x = -9 to x=−3x = -3: Total Area=2.5+5−6=1.5\text{Total Area} = 2.5 + 5 - 6 = 1.5
Answer
The definite integral ∫−9−3h(x) dx\int_{-9}^{-3} h(x) \, dx is 1.51.5.
Key Concept
Definite Integral of a Piecewise Linear Function
Explanation
To evaluate the definite integral of a piecewise linear function, break the integral into segments where the function is linear, calculate the area of each segment, and sum the areas.

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