   Chapter 4.3, Problem 42E

Chapter
Section
Textbook Problem

Using Properties of Definite Integrals Given ∫ 0 3 f ( x ) d x = 4     and     ∫ 3 6 f ( x ) d x = − 1 evaluate ∫ 0 6 f ( x ) d x ∫ 6 3 f ( x ) d x ∫ 3 3 f ( x ) d x ∫ 3 6 − 5 f ( x ) d x

(a)

To determine

To calculate: The definite integral 06f(x)dx using the values 03f(x)dx=436f(x)dx=1.

Explanation

Given:

The given values are

03f(x)dx=436f(x)dx=1

Formula used:

The additive interval property states that if f(x) is integrable on the three closed intervals determined by a,b and c, then abf(x)dx=acf(x)dx+cbf(x)dx

Calculation:

Evaluate the integral, that is, 06f(x)dx

As the given function is integrable on [0,6], it becomes

06

(b)

To determine

To calculate: The definite integral 63f(x)dx using the values 03f(x)dx=436f(x)dx=1.

(c)

To determine

To calculate: The definite integral 33f(x)dx using the values 03f(x)dx=436f(x)dx=1.

(d)

To determine

To calculate: The definite integral 365f(x)dx using the values 03f(x)dx=436f(x)dx=1.

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