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Posted 2 months ago

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Approximate the area between h(x)h(x) and the xx-axis from x=2x=-2 to x=4x=4 using a trapezoidal sum with 3 equal subdivisions.
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T(3)=

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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To approximate the area between h(x)h(x) and the xx-axis from x=2x = -2 to x=4x = 4 using a trapezoidal sum with 3 equal subdivisions, we first determine the width of each subdivision. The interval from 2-2 to 44 is 66 units long, so each subdivision has a width of 22 units
step 2
The trapezoidal rule formula for nn subdivisions is given by: T(n)=ba2n[h(x0)+2i=1n1h(xi)+h(xn)] T(n) = \frac{b - a}{2n} \left[ h(x_0) + 2 \sum_{i=1}^{n-1} h(x_i) + h(x_n) \right] For n=3n = 3, a=2a = -2, and b=4b = 4, the formula becomes: T(3)=4(2)23[h(2)+2(h(0)+h(2))+h(4)] T(3) = \frac{4 - (-2)}{2 \cdot 3} \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right]
step 3
Calculate the width of each subdivision: Δx=4(2)3=2 \Delta x = \frac{4 - (-2)}{3} = 2
step 4
Identify the xx values at the endpoints and midpoints of the subdivisions: x0=2,x1=0,x2=2,x3=4 x_0 = -2, \quad x_1 = 0, \quad x_2 = 2, \quad x_3 = 4
step 5
Substitute the xx values into the trapezoidal rule formula: T(3)=66[h(2)+2(h(0)+h(2))+h(4)] T(3) = \frac{6}{6} \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right] T(3)=1[h(2)+2(h(0)+h(2))+h(4)] T(3) = 1 \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right]
step 6
Using the given graph, approximate the values of h(x)h(x) at the specified points: h(2)0,h(0)4,h(2)6,h(4)2 h(-2) \approx 0, \quad h(0) \approx 4, \quad h(2) \approx 6, \quad h(4) \approx 2
step 7
Substitute these values into the formula: T(3)=1[0+2(4+6)+2] T(3) = 1 \left[ 0 + 2(4 + 6) + 2 \right] T(3)=1[0+2(10)+2] T(3) = 1 \left[ 0 + 2(10) + 2 \right] T(3)=1[0+20+2] T(3) = 1 \left[ 0 + 20 + 2 \right] T(3)=22 T(3) = 22
Answer
The approximate area between h(x)h(x) and the xx-axis from x=2x = -2 to x=4x = 4 using a trapezoidal sum with 3 equal subdivisions is 2222 square units.
Key Concept
Trapezoidal Rule
Explanation
The trapezoidal rule is a numerical method to approximate the definite integral of a function. It works by dividing the area under the curve into trapezoids, calculating the area of each, and summing them up.

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