Using trapezoidal rule, by dividing the interval [0, 4] into 4 equal parts, the approximate value of \[\int\limits_{0}^{4}{{{x}^{2}}+1}\] is equal to
(a) 25
(b) 26
(c) 27
(d) 28

Hint: Divide the interval into 4 parts thus find the sub interval of width \[\Delta x\], Now use the trapezoidal rule formula for 4 equal parts. Substitute x = 0, 1, 2, 3, 4 in f (x) get the values and substitute in the formula.
Complete step-by-step solution -
Trapezoidal rule is used for approximating the definite integrals where it uses the linear approximations of the function. Let f (x) be a continuous function on the interval [a, b] which is [0, 4]. Now divide the intervals [0, 4] into n equal subintervals with each of width, \[\Delta x\] i.e. n = 4.
\[\Delta x=\dfrac{b-a}{n}=\dfrac{4-0}{4}=\dfrac{4}{4}=1\]
Here, n = 4, as it is told to divide interval into 4 equal parts,
Here, \[f\left( x \right)=\int\limits_{0}^{4}{{{x}^{2}}+1}dx\]
Then the trapezoidal rule formula for area approximating the definite integral, \[\int\limits_{a}^{b}{f\left( x \right)}dx\] is given by,
\[\int\limits_{a}^{b}{f\left( x \right)}dx=\dfrac{\Delta x}{2}\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+.....+2f\left( {{x}_{n-1}} \right)+f\left( {{x}_{n}} \right) \right]\] where, \[{{x}_{i}}=a+i\Delta x\].
\[\int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=\dfrac{\Delta x}{2}\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+2f\left( {{x}_{3}} \right)+f\left( {{x}_{4}} \right) \right] \]...........(1)
Now let us find the values of \[f\left( {{x}_{0}} \right),f\left( {{x}_{1}} \right),f\left( {{x}_{2}} \right),f\left( {{x}_{3}} \right)\] and \[f\left( {{x}_{4}} \right)\], when x = 0,1,2,3,4 \[f\left( {{x}_{0}} \right)=f\left( 0 \right)={{x}^{2}}+1=0+1=1\].
\[\begin{align}
  & f\left( {{x}_{1}} \right)=f\left( 1 \right)={{x}^{2}}+1={{1}^{2}}+1=2 \\
 & f\left( {{x}_{2}} \right)=f\left( 2 \right)={{2}^{2}}+1=4+1=5 \\
 & f\left( {{x}_{3}} \right)=f\left( 3 \right)={{3}^{2}}+1=9+1=10 \\
 & f\left( {{x}_{4}} \right)=f\left( 4 \right)={{4}^{2}}+1=16+1=17 \\
\end{align}\]
Thus we got \[f\left( {{x}_{0}} \right)=1,f\left( {{x}_{1}} \right)=2,f\left( {{x}_{2}} \right)=5,f\left( {{x}_{3}} \right)=10\] and \[f\left( {{x}_{4}} \right)=17,\Delta x=1\].
Now let us substitute these values in equation (1).
\[\begin{align}
  & \int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=\dfrac{\Delta x}{2}\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+2f\left( {{x}_{3}} \right)+f\left( {{x}_{4}} \right) \right] \\
 & \int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=\dfrac{1}{2}\left[ 1+\left( 2\times 2 \right)+\left( 2\times 5 \right)+\left( 2\times 10 \right)+17 \right] \\
 & \int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=\dfrac{1}{2}\left[ 1+4+10+20+17 \right] \\
 & \int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=\dfrac{1}{2}\times 52=26 \\
\end{align}\]
Thus by dividing the interval [0, 4] into 4equal parts, the approximate value \[\int\limits_{0}^{4}{\left( {{x}^{2}}+1 \right)dx}=26\].
\[\therefore \] Option (b) is the correct answer.

Note: Trapezoidal rule integration works by approximating the region under the graph of a function as a trapezoid and calculating the area. If we compare trapezoidal rule to Simpson’s rule, trapezoidal rule doesn’t give accurate value, it is because trapezoidal rule uses linear approximations.