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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Area of a Region In Exercises 41 and 42, use integration to find the area of the triangular region having the given vertices.

( 0 , 0 ) , ( 6 , 0 ) , ( 4 , 3 )

To determine

To calculate: The area of the triangular region having the vertices (0,0), (6,0) and (4,3).

Explanation

Given Information:

The triangular region having the vertices (0,0), (6,0) and (4,3).

Formula used:

Area of a region bounded by two graphs is calculated using the following formula,

If f and g are continuous on [a,b] and g(x)f(x) for all x in [a,b], then the area of the region bounded by the graphs of f, g, x=a and x=b is given by

A=ab[f(x)g(x)]dx

The equation of a straight line passing through points (x1,y1) and (x2,y2) is,

yy1=(y2y1x2x1)(xx1)

Calculation:

Consider the triangular region having the vertices (0,0), (6,0) and (4,3). The triangle is shown as follows:

From the graph, triangle OAB is the required triangle. Its area will be equal to the sum of area below the function of side OA and the area below the function of side AB.

Compute the equation of side OA as follows:

Substitute (x1,y1)=(0,0) and (x2,y2)=(4,3) in the equation of a line, the equation of side OA is,

y0=(3040)(x0)y=34x

So, equation of line OA is y=34x

Compute the equation of side OB as follows:

Substitute (x1,y1)=(6,0) and (x2,y2)=(4,3) in the equation of a line, the equation of side OB is,

y0=(3046)(x6)

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