Chapter 5.5, Problem 1QQ

To determine

To choose : correct option for trapezoidal approximation of ${\displaystyle {\int }_1^{7}f\left(x\right)}dx$ .

1. $110$
2. $130$
3. $160$
4. $190$

Answer to Problem 1QQ

(C) is correct option.

Explanation of Solution

Given information :

The function $f$ is continuous on the closed interval $\left[1,7\right]$ and has values

  $\begin{array}{ccccc}x& 1& 4& 6& 7\\ f\left(x\right)& 10& 30& 40& 20\end{array}$

Formula needed :

Trapezoidal approximation ${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\frac{h}{2}\left(y_0+2y_1+2y_2+\cdot \cdot \cdot +2y_{h-1}+y_h\right)$ .

  ${\displaystyle {\int }_a^{b}f\left(x\right)}dx={\displaystyle {\int }_a^{t}f\left(x\right)}dx+{\displaystyle {\int }_t^{b}f\left(x\right)}dx$

Breaks the interval $\left[1,7\right]$ onto $\left[1,4\right],\left[4,6\right]$ and $\left[6,7\right]$ .

  ${\displaystyle {\int }_1^{7}f\left(x\right)}dx={\displaystyle {\int }_1^{4}f\left(x\right)}dx+{\displaystyle {\int }_4^{6}f\left(x\right)}dx+{\displaystyle {\int }_6^{7}f\left(x\right)}dx$

Apply trapezoidal approximation.

  ${\displaystyle {\int }_1^{7}f\left(x\right)}dx=\frac{h_1}{2}\left(f\left(1\right)+f\left(4\right)\right)+\frac{h_2}{2}\left(f\left(4\right)+f\left(6\right)\right)dx+\frac{h_3}{2}\left(f\left(6\right)+f\left(7\right)\right)$

Here, $h_1=4-1,h_2=6-4,h_3=7-6$ .

  $\begin{array}{c}{\displaystyle {\int }_1^{7}f\left(x\right)}dx=\frac{3}{2}\left(10+30\right)+\frac{2}{2}\left(30+40\right)+\frac{1}{2}\left(40+20\right)\\ =\frac{3}{2}\left(40\right)+\left(70\right)+\frac{1}{2}\left(60\right)\\ =3\cdot 20+70+30\\ =60+100\\ =160\end{array}$

The correct option is (C).