Question

How do you use the trapezoidal rule with \[n=4\] to approximate the area between the curve \[\sqrt{x}\sin x\] from \[\dfrac{\pi }{2}\] to \[\pi \]?

Answer 1

Hint: We are given the equation of a curve with \[n=4\] and we have to find the area in the boundary from \[\dfrac{\pi }{2}\] to \[\pi \] using the trapezoidal rule. So, first we will find the interval between the given limits, that is, finding the value of h, \[h=\dfrac{{{x}_{n}}-{{x}_{0}}}{n}\], where \[{{x}_{0}}=\dfrac{\pi }{2}\] and \[{{x}_{n}}=\pi \]. After obtaining the intervals, we will find the corresponding value of the curve given to us. And further we will use the trapezoidal rule, \[\int\limits_{{{x}_{0}}}^{{{x}_{n}}}{f(x)dx}=\dfrac{h}{2}[({{y}_{0}}+{{y}_{n}})+2({{y}_{1}}+{{y}_{2}}+...+{{y}_{n-1}})]\]. We will then have the value of the area bounded by the given curve.

Complete step by step solution:
According to the given question, we are given a curve \[\sqrt{x}\sin x\] with \[n=4\] and we have to find the area of the curve in the region from \[\dfrac{\pi }{2}\] to \[\pi \].
Let, \[\int\limits_{\dfrac{\pi }{2}}^{\pi }{\sqrt{x}\sin x}=\int\limits_{{{x}_{0}}}^{{{x}_{n}}}{f(x)dx}\]
We will first start with finding the interval of the given region and then we will find the corresponding value of the curve at those intervals.
So, we have,
\[h=\dfrac{{{x}_{n}}-{{x}_{0}}}{n}\]----(1)
Here, \[{{x}_{n}}=\pi \], \[n=4\] and \[{{x}_{0}}=\dfrac{\pi }{2}\],
Substituting the values in the equation (1), we get,
\[\Rightarrow h=\dfrac{\pi -\dfrac{\pi }{2}}{4}\]
Solving further we get,
\[\Rightarrow h=\dfrac{\dfrac{2\pi -\pi }{2}}{4}\]
\[\Rightarrow h=\dfrac{\pi }{8}\]
We have the value of the intervals, ‘h’ between the given regions. We will now obtain the corresponding value of these intervals using the equation of the curve in a tabular form. To obtain the next interval, we will add the value of ‘h’ to the previous interval in the given region. We have,

Value of ‘n’	Value of ‘x’	f(x)
0	\[\dfrac{\pi }{2}\]	1.253
1	\[\dfrac{5\pi }{8}\]	1.294
2	\[\dfrac{6\pi }{8}\]	1.085
3	\[\dfrac{7\pi }{8}\]	0.634
4	\[\pi \]	0.000

Now, we will use the trapezoidal rule to find the area, we have,
\[\int\limits_{{{x}_{0}}}^{{{x}_{n}}}{f(x)dx}=\dfrac{h}{2}[({{y}_{0}}+{{y}_{n}})+2({{y}_{1}}+{{y}_{2}}+...+{{y}_{n-1}})]\]
Using the above formula, we have the expression as,
\[\int\limits_{\dfrac{\pi }{2}}^{\pi }{\sqrt{x}\sin x}\]
\[\Rightarrow \dfrac{\dfrac{\pi }{8}}{2}[(1.253+0.000)+2(1.294+1.085+0.634)]\]
Solving the above expression, we have the solution as,
\[\Rightarrow \dfrac{\pi }{16}[(1.253)+2(3.014)]\]
Calculating it further, we have the value of the expression as,
\[\Rightarrow \dfrac{\pi }{16}[(1.253)+(6.028)]\]
\[\Rightarrow \dfrac{\pi }{16}[7.281]\]
\[\Rightarrow 1.429\]
Therefore, \[\int\limits_{\dfrac{\pi }{2}}^{\pi }{\sqrt{x}\sin x}=1.429\]

Note: While substituting the values in the formula of the trapezoid rule, it should be done correctly. And it is advised to do the calculations step wise. Also, the intervals should be carefully calculated else the entire answer will get wrong.