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Q32E

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Found in: Page 280

### Essential Calculus: Early Transcendentals

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

# 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 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.

Then,

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

Therefore,

The value of the integral is

$${\rm{ - 4}}{\rm{.5}}$$