Chapter 5, Problem 12RE

To determine

The value of the integral function.

Answer to Problem 12RE

The value of the integral is.$\underset{\_}{\frac{1}{10}}$.

Explanation of Solution

Given information:

The integral function is ${\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x\right)}^{9}dx}$.

The region lies between $x=0$ and $x=1$.

Calculation:

Consider $\left(1-x\right)$ as u,

$u=1-x$ (1)

Differentiate both side of the Equation.

$\begin{array}{c}du=\left(0-1\right)dx\\ =-dx\end{array}$ (2)

Calculate the lower limit value of u using Equation (1).

Substitute 0 for x in Equation (1).

$\begin{array}{c}u=1-0\\ =1.\end{array}$

Calculate the upper limit value of u using Equation (1).

Substitute 1 for x in Equation (1).

$\begin{array}{c}u=1-1\\ =0.\end{array}$

Substitute u for $\left(1-x\right)$ and $-du$ for dx in the function as shown below:

${\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x\right)}^{9}dx}={\displaystyle \underset{1}{\overset{0}{\int }}{u}^9\left(-du\right)}$ (2)

The expression to find the integral value using equation (2) is shown below:

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x\right)}^{9}dx}={\displaystyle \underset{1}{\overset{0}{\int }}{u}^9\left(-du\right)}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{u}^9\left(du\right)}\\ ={\left(\frac{u^{10}}{10}\right)}_0^1\\ =\left(\frac{1^{10}}{10}-\frac{{\left(0\right)}^{10}}{10}\right)\end{array}$

${\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x\right)}^{9}dx}=\frac{1}{10}$

Therefore, the value of the integral is $\underset{\_}{\frac{1}{10}}$.