Question

How do you use the trapezoidal rule with \[n=6\] to approximate the integral \[\int\limits_{0}^{3}{\dfrac{dx}{1+{{x}^{2}}+{{x}^{4}}}}\] ?

Answer 1

Hint: Problems like these are a bit tricky but are easy to solve once we know the concept behind the problem and what all formulae and theorems would be required for solving it. In trapezoidal rule for integral approximation, we first of all need to divide the given interval within which we are needed to integrate the function into ‘n’ equal subintervals. In our problem, the given interval is from \[\left[ 0,3 \right]\] and we need to divide this interval into \[n=6\] equal subparts. After dividing the interval into subparts, we need to apply the formula for approximation by trapezoidal rule. We define any integral as,
\[\int\limits_{a}^{b}{f\left( x \right)dx}\] . We divide this as, \[\left[ 0,0.5 \right],\left[ 0.5,1 \right],\left[ 1,1.5 \right],\left[ 1.5,2 \right],\left[ 2,2.5 \right],\left[ 2.5,3 \right]\]

Complete step-by-step solution:
Now the formula to find the approximate value is \[{{T}_{n}}=\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+2f\left( {{x}_{3}} \right)+2f\left( {{x}_{4}} \right)+2f\left( {{x}_{5}} \right)+f\left( {{x}_{6}} \right) \right]\left( \dfrac{b-a}{2\times 6} \right)\]

Now we start off with the solution the given problem by writing the formula of the Trapezoidal rule as,
\[{{T}_{n}}=\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+2f\left( {{x}_{3}} \right)+2f\left( {{x}_{4}} \right)+2f\left( {{x}_{5}} \right)+f\left( {{x}_{6}} \right) \right]\left( \dfrac{b-a}{2\times 6} \right)\]
In this formula we put the value of the unknown parameters as,
\[\Rightarrow {{T}_{n}}=\left[ f\left( 0 \right)+2f\left( 0.5 \right)+2f\left( 1 \right)+2f\left( 1.5 \right)+2f\left( 2 \right)+2f\left( 2.5 \right)+f\left( 3 \right) \right]\left( \dfrac{3}{2\times 6} \right)\]
Now if we take a closer look at our problem, then we can clearly see that the value of the function which is taken in this problem can be written down as,
\[\Rightarrow f\left( x \right)=\dfrac{1}{1+{{x}^{2}}+{{x}^{4}}}\]
Now we are required to find out the values of the respective points as,
\[f\left( 0 \right)=1\] , \[f\left( 0.5 \right)=.7619\] , \[f\left( 1 \right)=.3333\] , \[f\left( 1.5 \right)=.1203\] , \[f\left( 2 \right)=.0476\] , \[f\left( 2.5 \right)=.0216\] , \[f\left( 3 \right)=.0110\]
Now we put these value into our formula to get the required value of the integral as,
\[\Rightarrow {{T}_{n}}=\left[ 1+2\times .7619+2\times .3333+2\times .1203+2\times .0476+2\times .0216+.0110 \right]\left( \dfrac{1}{4} \right)\]
Evaluating the value we get,
\[\begin{align}
  & \Rightarrow {{T}_{n}}=\left[ 1+1.5238+.6666+.3406+0.0952+.0432+.0110 \right]\left( \dfrac{1}{4} \right) \\
 & \Rightarrow {{T}_{n}}=\left[ 3.6804 \right]\left( \dfrac{1}{4} \right) \\
 & \Rightarrow {{T}_{n}}=0.9201 \\
\end{align}\]
Thus, our value of the integral is \[.9201\] . However, this value is an approximate one.

Note: We need to know the formulae for the trapezoidal rule and some of the basics of integration and differentiation. We must be very careful while dividing the given interval into ‘n’ given parts equally, as if it is done in a wrong way may yield wrong results. We must also note one very important thing for these types of sums, that these are just approximations and may vary with the actual area under the curve. To get a closer and accurate answer we need to divide the interval into a greater number of parts.