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### Essential Calculus: Early Transcendentals

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

# 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 $$3.$$

See the step by step solution

## 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

4.

## 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.