Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

In Example 2 in Section 5.1 we showed that \(\int_0^1 {{x^2}} dx = \frac{1}{3}\). Use this fact and the properties of integrals to evaluate \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\).

The value of \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\) is \(3.\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Property Used for the integral

Suppose all the following integrals exist.

1. \(\int_a^b c dx = c(b - a)\), where \(c\) is any constant

2. \(\int_a^b {(f(x) + g(x))dx} = \int_a^b {f(x)dx} + \int_a^b {g(x)dx} \)

3. \(\int_a^b c f(x)dx = c\int_a^b f (x)dx\), where \(c\) is any constant

Step 2: Obtain the value of the integral

The value of the integral, \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\) as follows.

Apply the properties 3 and 4 in the integral \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\) and compute.

\(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx = \int_0^1 5 dx - \int_0^1 6 {x^2}dx\)

\( = \int_0^1 5 dx - 6\int_0^1 {{x^2}} dx{\kern 1pt} \,{\kern 1pt} {\kern 1pt} \,........(1)\)

Apply property 1 in \(\int_0^1 5 dx\) and calculate as,

\(\int_0^1 5 dx = 5(1 - 0)\)

\( = 5\)

Step 3: Substitute the values in the given integral

It is given that, \(\int_0^1 {{x^2}} dx = \frac{1}{3}\). Substitute this in (1) and compute the value of the integral as follows.

\(\begin{aligned}{l}\int_0^1 {\left( {5 - 6{x^2}} \right)} dx &= \int_0^1 5 dx - 6\int_0^1 {{x^2}} dx\\ &= 5 - 6 \times \frac{1}{3}\\ &= 5 - 2\\ &= 3\end{aligned}\)

Hence, the value of \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\) is 3.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

In Example 2 in Section 5.1 we showed that \(\int_0^1 {{x^2}} dx = \frac{1}{3}\). Use this fact and the properties of integrals to evaluate \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\).

Step by Step Solution

Step 1: Property Used for the integral

Step 2: Obtain the value of the integral

Step 3: Substitute the values in the given integral

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

In Example 2 in Section 5.1 we showed that \(\int_0^1 {{x^2}} dx = \frac{1}{3}\). Use this fact and the properties of integrals to evaluate \(\int_0^1 {\left( {5 - 6{x^2}} \right)} dx\).

Step by Step Solution

Step 1: Property Used for the integral

Step 2: Obtain the value of the integral

Step 3: Substitute the values in the given integral

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