Question

3 during the time interval ( 0 leq t leq 8 ) minutes, water drains from a large vat at a rate of ( g(t) ) gallons/minute. the function ( g(t) ) is a continuous function with select values indicated in the table. use two trapezoids of equal width to approximate the amount of water (in gallons) that drains from the vat over the time interval (0,8).

( t ) (minutes)	0	1	2	3	4	5	6	7	8
( g(t) ) (gal/min)	2	3	3	5	4	1	0	2	5

Explanation:

To approximate the amount of water drained using two trapezoids of equal width over the interval \([0, 8]\), we first determine the width of each trapezoid. The total interval length is \(8 - 0 = 8\) minutes, and we are using 2 trapezoids, so the width \(\Delta t\) of each trapezoid is \(\frac{8 - 0}{2} = 4\) minutes.

Step 1: Define the intervals and corresponding function values

The first trapezoid is over the interval \([0, 4]\) with \(t = 0, 1, 2, 3, 4\) and the second trapezoid is over the interval \([4, 8]\) with \(t = 4, 5, 6, 7, 8\). But since we are using trapezoidal approximation with two trapezoids, we can also consider the endpoints of each sub - interval. For the trapezoidal rule, the formula for the area of a trapezoid with bases \(b_1\) and \(b_2\) and height \(h\) is \(A=\frac{(b_1 + b_2)h}{2}\).

For the first trapezoid (interval \([0,4]\)):
The "bases" of the trapezoid (the values of \(g(t)\) at the endpoints) are \(g(0) = 2\) and \(g(4)=4\), and the width (height of the trapezoid in terms of the \(t\) - axis) \(\Delta t_1=4 - 0 = 4\). The area of the first trapezoid \(A_1=\frac{(g(0)+g(4))\times\Delta t_1}{2}=\frac{(2 + 4)\times4}{2}\)

For the second trapezoid (interval \([4,8]\)):
The "bases" are \(g(4) = 4\) and \(g(8)=5\), and the width \(\Delta t_2=8 - 4 = 4\). The area of the second trapezoid \(A_2=\frac{(g(4)+g(8))\times\Delta t_2}{2}=\frac{(4 + 5)\times4}{2}\)

Step 2: Calculate the area of each trapezoid

- Calculate \(A_1\):
\(A_1=\frac{(2 + 4)\times4}{2}=\frac{6\times4}{2}=12\)
- Calculate \(A_2\):
\(A_2=\frac{(4 + 5)\times4}{2}=\frac{9\times4}{2}=18\)

Step 3: Calculate the total amount of water drained

The total amount of water drained \(A\) is the sum of the areas of the two trapezoids. So \(A=A_1 + A_2\)
\(A=12 + 18=30\)

Wait, we can also use the trapezoidal rule with more sub - intervals within each of the two main trapezoids. Let's re - evaluate. If we consider the two trapezoids of equal width (\(\Delta t = 4\)), but we can also use the trapezoidal rule on the sub - intervals. Wait, the problem says "use two trapezoids of equal width". The interval \([0,8]\) is divided into two sub - intervals: \([0,4]\) and \([4,8]\), each with width \(\Delta t=4\).

But let's check the values again. The function values are:
At \(t = 0\), \(g(0)=2\); \(t = 4\), \(g(4) = 4\); \(t = 8\), \(g(8)=5\)

But maybe we should use the trapezoidal rule on the sub - intervals within \([0,4]\) and \([4,8]\) as trapezoids. Wait, the trapezoidal rule for a function \(y = f(x)\) over \([a,b]\) with \(n\) sub - intervals is \(\int_{a}^{b}f(x)dx\approx\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]\). But here we are told to use two trapezoids of equal width. So the width of each trapezoid is \(4\).

First trapezoid: from \(t = 0\) to \(t = 4\). The trapezoidal approximation for \(\int_{0}^{4}g(t)dt\) can also be calculated by considering the trapezoids between each pair of points. The points in \([0,4]\) are \(t = 0,1,2,3,4\) with \(g(0)=2\), \(g(1)=3\), \(g(2)=3\), \(g(3)=5\), \(g(4)=4\). The width between each \(t\) - value is \(\Delta t = 1\). The trapezoidal rule for \([0,4]\) with \(n = 4\) sub - intervals (width \(\Delta t=1\)) is \(\frac{1}{2}[g(0)+2g(1)+2g(2)+2g(3)+g(4)]=\frac{1}{2}[2 + 2\times3+2\times3 + 2\times5+4]=\frac{1}{2}[2 + 6+6 + 10+4]=\frac{1}{2}[28]=14\)

Second trapezoid: from \(t = 4\) to \(t = 8\). The points are \(t = 4,5,6,7,8\) with \(g(4)=4\), \(g(5)=1\), \(g(6)=0\), \(g(7)=2\), \(g(8)=5\). Using the trapezoidal rule with \(n = 4\) sub - intervals (width \(\Delta t = 1\)): \(\frac{1}{2}[g(4)+2g(5)+2g(6)+2g(7)+g(8)]=\frac{1}{2}[4+2\times1 + 2\times0+2\times2+5]=\frac{1}{2}[4 + 2+0 + 4+5]=\frac{1}{2}[15]=7.5\)

Wait, but the problem says "use two trapezoids of equal width". So the width of each trapezoid is \(4\) (since \(8\div2 = 4\)). So the first trapezoid is from \(t = 0\) to \(t = 4\), and the second from \(t = 4\) to \(t = 8\).

The formula for the area of a trapezoid is \(A=\frac{(b_1 + b_2)h}{2}\), where \(b_1\) and \(b_2\) are the lengths of the two parallel sides (the values of \(g(t)\) at the endpoints of the interval) and \(h\) is the distance between them (the width of the interval).

For the first trapezoid (\([0,4]\)):
We can also calculate the area by summing the areas of the trapezoids between each \(t\) - value. From \(t = 0\) to \(t = 1\): \(A_{0 - 1}=\frac{(g(0)+g(1))\times1}{2}=\frac{(2 + 3)\times1}{2}=2.5\)
From \(t = 1\) to \(t = 2\): \(A_{1 - 2}=\frac{(g(1)+g(2))\times1}{2}=\frac{(3 + 3)\times1}{2}=3\)
From \(t = 2\) to \(t = 3\): \(A_{2 - 3}=\frac{(g(2)+g(3))\times1}{2}=\frac{(3 + 5)\times1}{2}=4\)
From \(t = 3\) to \(t = 4\): \(A_{3 - 4}=\frac{(g(3)+g(4))\times1}{2}=\frac{(5 + 4)\times1}{2}=4.5\)
Sum of areas from \(0\) to \(4\): \(2.5+3 + 4+4.5 = 14\)

For the second trapezoid (\([4,8]\)):
From \(t = 4\) to \(t = 5\): \(A_{4 - 5}=\frac{(g(4)+g(5))\times1}{2}=\frac{(4 + 1)\times1}{2}=2.5\)
From \(t = 5\) to \(t = 6\): \(A_{5 - 6}=\frac{(g(5)+g(6))\times1}{2}=\frac{(1 + 0)\times1}{2}=0.5\)
From \(t = 6\) to \(t = 7\): \(A_{6 - 7}=\frac{(g(6)+g(7))\times1}{2}=\frac{(0 + 2)\times1}{2}=1\)
From \(t = 7\) to \(t = 8\): \(A_{7 - 8}=\frac{(g(7)+g(8))\times1}{2}=\frac{(2 + 5)\times1}{2}=3.5\)
Sum of areas from \(4\) to \(8\): \(2.5+0.5 + 1+3.5 = 7.5\)

Total amount of water drained: \(14 + 7.5=21.5\)? Wait, no, the problem says "use two trapezoids of equal width". So the width of each trapezoid is \(4\). So we should consider the two trapezoids as:

First trapezoid: endpoints at \(t = 0\) and \(t = 4\), with the "average" of the function values? No, the trapezoidal rule for two trapezoids over \([0,8]\) means that we divide \([0,8]\) into two sub - intervals: \([0,4]\) and \([4,8]\), each with \(\Delta t = 4\).

The formula for the trapezoidal approximation with \(n = 2\) sub - intervals is:

\(\int_{0}^{8}g(t)dt\approx\frac{\Delta t}{2}[g(0)+2g(4)+g(8)]\) (Wait, no, the trapezoidal rule with \(n\) sub - intervals has the formula \(\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]\). When \(n = 2\), \(\Delta x=\frac{8 - 0}{2}=4\), \(x_0 = 0\), \(x_1 = 4\), \(x_2 = 8\). So the formula is \(\frac{4}{2}[g(0)+2g(4)+g(8)]=2[2 + 2\times4+5]=2[2 + 8 + 5]=2\times15 = 30\)

But when we calculate the sum of the trapezoids with \(\Delta t = 1\) (using all the given points), we get a different result. The confusion comes from the interpretation of "two trapezoids of equal width". If we consider each trapezoid as a single trapezoid with width \(4\), with bases \(g(0)\) and \(g(4)\) for the first, and \(g(4)\) and \(g(8)\) for the second, then:

Area of first trapezoid: \(\frac{(g(0)+g(4))\times4}{2}=\frac{(2 + 4)\times4}{2}=12\)

Area of second trapezoid: \(\frac{(g(4)+g(8))\times4}{2}=\frac{(4 + 5)\times4}{2}=18\)

Total: \(12 + 18 = 30\)

If we consider the trapezoids with the intermediate points, we are using more than two trapezoids. Since the problem says "two trapezoids of equal width", we should divide the interval \([0,8]\) into two parts of width \(4\) each, and approximate the integral over each part as a trapezoid with bases at the endpoints of the part.

Answer:

\(30\)