   Chapter 4.3, Problem 56E

Chapter
Section
Textbook Problem

Verifying a Property Use a graph to explain why ∫ a b k f ( x ) d x = k ∫ a b f ( x )   d x if f is integrable on [a, b] and k is a constant.

To determine

To prove: The definite integrals abkf(x)dx=kabf(x)dx.

Explanation

Given: The property to be verified is,

abkf(x)dx=kabf(x)dx

Where f(x) is integrable on [a,b] and k is a constant.

Formula used: We know that area bounded by the graph of f(x) and x -axis on interval [a,b] is represented by,

abf(x)dx

Proof: Take a function f(x)=3x which is integrable on [1,2]

First we will draw the graph of f(x)=3x for that substitute x=1,2 in f(x).

At x=1,

f(1)=3(1)=3

And,

At x=2

f(2)=3(2)=6

Any line can be drawn using two points and the points are (1,3),(2,6) plot the points on the coordinate axes and shade the region to obtain,

This figure is a trapezoid of altitude 1 and the parallel bases of 3 and 6.

So,

12(3x)dx=area of trapezoid.

Using, (area of trapezoid)=12h(a+b) to get,

12(3x)dx=12(1)(3+6)=92

Now, draw the graph of f(x)=x for that substitute x=1,2 in f(x)

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