Question

A river is 80 meters wide. The depth d in meters at a distance x meter from one bank to the other bank is given by the following table

x meters	0	10	20	30	40	50	60	70	80
d meters	0	4	7	9	12	15	14	8	0

The approximate area of the cross section using Simpson’s rule is
(a) 700 sq. Meters
(b) 800 sq. Meters
(c) 720 sq. Meters
(d) 740 sq. Meters

Answer 1

Hint : To solve the given question, we will use the Simpson’s one – third rile of integration which says that the integration of a function f(x) is approximated by a second-order polynomial. It is defined by:
$\int\limits_{a}^{b}{f\left( x \right)}dx=\dfrac{h}{3}\left[ \left( {{y}_{0}}+{{y}_{n}} \right)+4\left( {{y}_{1}}+{{y}_{3}}+{{y}_{5}}+.....+{{y}_{n-1}} \right)+2\left( {{y}_{2}}+{{y}_{4}}+....+{{y}_{n-2}} \right) \right]$
where h is the difference between the terms and ${{y}_{0}},{{y}_{1}},......{{y}_{n}}$ are equally spaced. So, we will find out the value of h and ${{y}_{0}},{{y}_{1}},......{{y}_{n}}$ and then we will put these in the above formula to get the appropriate area.

Complete step by step solution :
Before we solve the given question, we will first find out what Simpson’s rule is. Simpson’s rule is a numerical method that is used to evaluate the definite integral. Let us assume that the depth of the river is a function of distance x, i.e. d = f(x). Now, we have to calculate the area of the river. We know that the area of any function g(x) over an interval [a, b] is given by
$\text{Area}=\int\limits_{a}^{b}{g\left( x \right)dx}$
Now, according to Simpson’s one – third method, the integral of a definite function over an interval is given by
$\int\limits_{a}^{b}{g\left( x \right)}dx=\dfrac{h}{3}\left[ \left( {{y}_{0}}+{{y}_{n}} \right)+4\left( {{y}_{1}}+{{y}_{3}}+{{y}_{5}}+.....+{{y}_{n-1}} \right)+2\left( {{y}_{2}}+{{y}_{4}}+....+{{y}_{n-2}} \right) \right]$
where $h=\dfrac{b-a}{n}$ and n is the number of divisions and ${{y}_{0}},{{y}_{1}},.....{{y}_{n}}$ are the terms which are equally spaced between a and b.
In our case, the river is 80 meters long, so a = 0 and b = 80. There are total of 8 divisions given in the question. Thus,
$h=\dfrac{80-0}{8}$
$\Rightarrow h=\dfrac{80}{8}$
$\Rightarrow h=10$
The values of ${{y}_{0}},{{y}_{1}},.....{{y}_{8}}$ are given in the question. Thus, the area will be,
$\text{Area}=\int{\left( d \right)dx}=\int{f\left( x \right)dx}$
$\Rightarrow \text{Area}=\dfrac{10}{3}\left[ \left( 0+0 \right)+4\left( 4+9+15+8 \right)+2\left( 7+12+14 \right) \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{10}{3}\left[ 0+4\left( 36 \right)+2\left( 33 \right) \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{10}{3}\left[ 144+66 \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{10}{3}\left[ 210 \right]sq.meters$
$\Rightarrow \text{Area}=700sq.meters$
Hence, option (b) is correct.

Note : In the above solution, we have used Simpson’s one – third rule of integration. In place of this, we can all use Simpson’s $\dfrac{3}{8}$ rule of integration. According to this, we have,
$\text{Area}=\dfrac{3h}{8}\left[ \left( {{y}_{0}}+{{y}_{8}} \right)+3\left( {{y}_{1}}+{{y}_{2}}+{{y}_{4}}+{{y}_{5}}+{{y}_{7}} \right)+2\left( {{y}_{3}}+{{y}_{6}} \right) \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{3\times 10}{8}\left[ \left( 0+0 \right)+3\left( 4+7+12+15+8 \right)+2\left( 9+14 \right) \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{15}{4}\left[ 0+3\left( 46 \right)+2\left( 23 \right) \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{15}{4}\left[ 138+46 \right]sq.meters$
$\Rightarrow \text{Area}=\dfrac{15}{4}\left[ 184 \right]sq.meters$
$\Rightarrow \text{Area}=15\times 46\text{ }sq.meters$
$\Rightarrow \text{Area}=690\text{ }sq.meters$
$\Rightarrow \text{Area}\simeq 700\text{ }sq.meters$
One thing to note is that these are not the actual areas. These are just approximated areas, so they may vary.