Chapter 6.5, Problem 4QQ

To determine

Trapezoidal approximation of the integral.

Answer to Problem 4QQ

Trapezoidal approximation:

  ${\displaystyle {\int }_1^{7}f\left(x\right)dx}\approx 160$

Explanation of Solution

Given information:

Function f is continuous on the closed interval [1, 7].

Function values:

  

The widths of subinterval:

[1, 4], [4, 6], and [6, 7].

Since the widths of the subintervals are not same, the Trap Rule formula cannot be used directly.

Instead, find the area of each trapezoid and then sum them up.

For [1, 4]:

Width,

  $b=4-1=3$

Heights,

  $h_1=10$

And

  $h_2=30$

Then

The area for this trapezoid,

  $A_1=\frac{1}{2}b\left(h_1+h_2\right)=\frac{1}{2}\left(3\right)\left(10+30\right)=\frac{3}{2}\left(40\right)=60$

For [4, 6]:

Width,

  $b=6-4=2$

Heights,

  $h_1=30$

And

  $h_2=40$

Then

The area for this trapezoid,

  $A_2=\frac{1}{2}b\left(h_1+h_2\right)=\frac{1}{2}\left(2\right)\left(30+40\right)=\left(1\right)\left(70\right)=70$

For [6, 7]:

Width,

  $b=7-6=1$

Heights,

  $h_1=40$

And

  $h_2=20$

Then

The area for this trapezoid,

  $A_3=\frac{1}{2}b\left(h_1+h_2\right)=\frac{1}{2}\left(1\right)\left(40+20\right)=\frac{1}{2}\left(60\right)=30$

Thus,

The approximation of integral:

  ${\displaystyle {\int }_1^{7}f\left(x\right)dx}\approx A_1+A_2+A_3\approx 60+70+30=160$