Question

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

Answer 1

Hint: Divide the interval into 4 parts thus find the sub interval of width $$\mathrm{\Delta }x$$, 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, $$\mathrm{\Delta }x$$ i.e. n = 4.
$$\mathrm{\Delta }x={\displaystyle \frac{b-a}{n}}={\displaystyle \frac{4-0}{4}}={\displaystyle \frac{4}{4}}=1$$
Here, n = 4, as it is told to divide interval into 4 equal parts,
Here, $$f\left(x\right)=\underset{0}{\overset{4}{\int }}{x}^2+1dx$$
Then the trapezoidal rule formula for area approximating the definite integral, $$\underset{a}{\overset{b}{\int }}f\left(x\right)dx$$ is given by,
$$\underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \frac{\mathrm{\Delta }x}{2}}[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+.....+2f\left(x_{n-1}\right)+f\left(x_n\right)]$$ where, $$x_i=a+i\mathrm{\Delta }x$$.
$$\underset{0}{\overset{4}{\int }}(x^2+1)dx={\displaystyle \frac{\mathrm{\Delta }x}{2}}[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_3\right)+f\left(x_4\right)]$$...........(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 $$f\left(x_0\right)=f\left(0\right)=x^2+1=0+1=1$$.
$$\begin{array}{rl}& f\left(x_1\right)=f\left(1\right)=x^2+1=1^2+1=2\\ & f\left(x_2\right)=f\left(2\right)=2^2+1=4+1=5\\ & f\left(x_3\right)=f\left(3\right)=3^2+1=9+1=10\\ & f\left(x_4\right)=f\left(4\right)=4^2+1=16+1=17\end{array}$$
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 $$f\left(x_4\right)=17,\mathrm{\Delta }x=1$$.
Now let us substitute these values in equation (1).
$$\begin{array}{rl}& \underset{0}{\overset{4}{\int }}(x^2+1)dx={\displaystyle \frac{\mathrm{\Delta }x}{2}}[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_3\right)+f\left(x_4\right)]\\ & \underset{0}{\overset{4}{\int }}(x^2+1)dx={\displaystyle \frac{1}{2}}[1+(2\times 2)+(2\times 5)+(2\times 10)+17]\\ & \underset{0}{\overset{4}{\int }}(x^2+1)dx={\displaystyle \frac{1}{2}}[1+4+10+20+17]\\ & \underset{0}{\overset{4}{\int }}(x^2+1)dx={\displaystyle \frac{1}{2}}\times 52=26\end{array}$$
Thus by dividing the interval [0, 4] into 4equal parts, the approximate value $$\underset{0}{\overset{4}{\int }}(x^2+1)dx=26$$.
$$\therefore$$ 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.