Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 280
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Evaluate the integral by interpreting it in terms of areas

\(\int\limits_{\rm{0}}^{\rm{9}} {\left( {\frac{{\rm{1}}}{{\rm{3}}}{\rm{x - 2}}} \right){\rm{dx}}} {\rm{.}}\)

The value of integral is \({\rm{ - 4}}{\rm{.5}}\)

See the step by step solution

Step by Step Solution

Step 1: Objective of the question.

Consider the following integral:

\(\int\limits_{\rm{0}}^{\rm{9}} {\left( {\frac{{\rm{1}}}{{\rm{3}}}{\rm{x - 2}}} \right){\rm{dx}}} \)

The objective is to interpret the definite integral \(\int\limits_{\rm{0}}^{\rm{9}} {\left( {\frac{{\rm{1}}}{{\rm{3}}}{\rm{x - 2}}} \right){\rm{dx}}} \) is as the area of the region bounded between the function \({\rm{f(x) = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{x - 2}}\) and the axis over the interval \(\left( {{\rm{0,9}}} \right)\)

The region is as shown below:

Step 2: Given information.

The shaded region consists of two triangles.

Region A is a triangle with base length \({\rm{6}}\)and height \({\rm{2}}\)

So, the area of region A is as follow:

\({\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{bh = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(6)(2) = 6}}\)

Step 3: Area of Triangle.

Region B is a triangle with base length \({\rm{3}}\) and height \({\rm{1}}\)

So, the area of the region B is as follow:

\({\rm{B = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{bh = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(3)(1) = 1}}{\rm{.5}}\)

Step 4: Find the total integral.

The total integral is the sum of the positive areas above the \({\rm{x}}\)axis and the negative area below the \({\rm{x}}\)axis.


\(\int\limits_{\rm{0}}^{\rm{9}} {\left( {\frac{{\rm{1}}}{{\rm{3}}}{\rm{x - 2}}} \right){\rm{dx = - A + B}}} {\rm{ = - 6 + 1}}{\rm{.5 = - 4}}{\rm{.5}}\)


The value of the integral is

\({\rm{ - 4}}{\rm{.5}}\)

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.