To solve this problem, we will use Simpson's rule, which is a method for numerical integration. The formula for Simpson's rule is:

∫(a to b) f(x) dx ≈ (b - a) / 6 * [f(a) + 4f((a+b)/2) + f(b)]

The rate of flow of the river is given by the cross-sectional area multiplied by the velocity of the river. The cross-sectional area can be approximated using Simpson's rule.

The given depths at 1m intervals are: 0, 1.5, 2, 2.8, 2.8, 1.9, 0.

We can divide these into three sections: [0, 1.5, 2], [2, 2.8, 2.8], and [2.8, 1.9, 0].

Applying Simpson's rule to each section:

Area1 = 1/6 * [f(0) + 4f(1.5) + f(2)] = 1/6 * [0 + 4*1.5 + 2] = 1.5 m^2
Area2 = 1/6 * [f(2) + 4f(2.8) + f(2.8)] = 1/6 * [2 + 4*2.8 + 2.8] = 3.2 m^2
Area3 = 1/6 * [f(2.8) + 4f(1.9) + f(0)] = 1/6 * [2.8 + 4*1.9 + 0] = 2.3 m^2

The total area is the sum of these, which is 1.5 + 3.2 + 2.3 = 7 m^2.

The rate of flow of the river is then the total area multiplied by the velocity, which is 7 m^2 * 2 m/s = 14 m^3/s.

So, the approximate rate of flow of the river is 14 cubic meters per second.

Question

To solve this problem, we will use Simpson's rule, which is a method for numerical integration. The formula for Simpson's rule is:

∫(a to b) f(x) dx ≈ (b - a) / 6 * [f(a) + 4f((a+b)/2) + f(b)]

The rate of flow of the river is given by the cross-sectional area multiplied by the velocity of the river. The cross-sectional area can be approximated using Simpson's rule.

The given depths at 1m intervals are: 0, 1.5, 2, 2.8, 2.8, 1.9, 0.

We can divide these into three sections: [0, 1.5, 2], [2, 2.8, 2.8], and [2.8, 1.9, 0].

Applying Simpson's rule to each section:

Area1 = 1/6 * [f(0) + 4f(1.5) + f(2)] = 1/6 * [0 + 4*1.5 + 2] = 1.5 m^2
Area2 = 1/6 * [f(2) + 4f(2.8) + f(2.8)] = 1/6 * [2 + 4*2.8 + 2.8] = 3.2 m^2
Area3 = 1/6 * [f(2.8) + 4f(1.9) + f(0)] = 1/6 * [2.8 + 4*1.9 + 0] = 2.3 m^2

The total area is the sum of these, which is 1.5 + 3.2 + 2.3 = 7 m^2.

The rate of flow of the river is then the total area multiplied by the velocity, which is 7 m^2 * 2 m/s = 14 m^3/s.

So, the approximate rate of flow of the river is 14 cubic meters per second.

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To solve this problem, we will use Simpson's rule, which is a method for numerical integration. The formula for Simpson's rule is:

∫(a to b) f(x) dx ≈ (b - a) / 6 * [f(a) + 4f((a+b)/2) + f(b)]

The rate of flow of the river is given by the cross-sectional area multiplied by the velocity of the river. The cross-sectional area can be approximated using Simpson's rule.

The given depths at 1m intervals are: 0, 1.5, 2, 2.8, 2.8, 1.9, 0.

We can divide these into three sections: [0, 1.5, 2], [2, 2.8, 2.8], and [2.8, 1.9, 0].

Applying Simpson's rule to each section:

Area1 = 1/6 * [f(0) + 4f(1.5) + f(2)] = 1/6 * [0 + 4*1.5 + 2] = 1.5 m^2
Area2 = 1/6 * [f(2) + 4f(2.8) + f(2.8)] = 1/6 * [2 + 4*2.8 + 2.8] = 3.2 m^2
Area3 = 1/6 * [f(2.8) + 4f(1.9) + f(0)] = 1/6 * [2.8 + 4*1.9 + 0] = 2.3 m^2

The total area is the sum of these, which is 1.5 + 3.2 + 2.3 = 7 m^2.

The rate of flow of the river is then the total area multiplied by the velocity, which is 7 m^2 * 2 m/s = 14 m^3/s.

So, the approximate rate of flow of the river is 14 cubic meters per second.

A river is 6 metres wide and is flowing at a uniform velocity of 2𝑚/𝑠. The depth of a crosssection of the river, measured at 1𝑚 intervals, is given in metres as:0 1.5 2 2.8 2.8 1.9 0Use Simpson’s rule to approximate the rate of flow of the river, that is, the rate of change ofvolume

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