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Math

Posted 2 months ago

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

Posted 2 months ago

Solution by Steps

step 1

To approximate the area between $h(x)$ and the $x$-axis from $x = -2$ to $x = 4$ using a trapezoidal sum with 3 equal subdivisions, we first determine the width of each subdivision. The interval from $-2$ to $4$ is $6$ units long, so each subdivision has a width of $2$ units

step 2

The trapezoidal rule formula for $n$ subdivisions is given by:
$T(n) = \frac{b - a}{2n} \left[ h(x_0) + 2 \sum_{i=1}^{n-1} h(x_i) + h(x_n) \right]$
For $n = 3$, $a = -2$, and $b = 4$, the formula becomes:
$T(3) = \frac{4 - (-2)}{2 \cdot 3} \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right]$

step 3

Calculate the width of each subdivision:
$\Delta x = \frac{4 - (-2)}{3} = 2$

step 4

Identify the $x$ values at the endpoints and midpoints of the subdivisions:
$x_0 = -2, \quad x_1 = 0, \quad x_2 = 2, \quad x_3 = 4$

step 5

Substitute the $x$ values into the trapezoidal rule formula:
$T(3) = \frac{6}{6} \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right]$
$T(3) = 1 \left[ h(-2) + 2(h(0) + h(2)) + h(4) \right]$

step 6

Using the given graph, approximate the values of $h(x)$ at the specified points:
$h(-2) \approx 0, \quad h(0) \approx 4, \quad h(2) \approx 6, \quad h(4) \approx 2$

step 7

Substitute these values into the formula:
$T(3) = 1 \left[ 0 + 2(4 + 6) + 2 \right]$
$T(3) = 1 \left[ 0 + 2(10) + 2 \right]$
$T(3) = 1 \left[ 0 + 20 + 2 \right]$
$T(3) = 22$

Answer

The approximate area between $h(x)$ and the $x$-axis from $x = -2$ to $x = 4$ using a trapezoidal sum with 3 equal subdivisions is $22$ 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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