Chapter 6.3, Problem 34E

To determine

To find: The area of the region enclosed by the $y$ -axis .

Answer to Problem 34E

The area is: $0.726$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{c}y=x^2\\ y=\left(x^2+x+1\right)e^{-x}\end{array}$

Calculation:

The graph of the functions

  $\begin{array}{c}y=x^2\\ y=\left(x^2+x+1\right)e^{-x}\end{array}$

on a graphing calculator to find their intersection. Since the region is bounded on the left by the $y$ -axis and the intersection point $x\approx 1.05$ the area of the region is bounded by

  $\begin{array}{c}x=0\\ x=1.05\end{array}$

The area of the region is then the area under $y=\left(x^2+\right.x+1)e^{-x}$ minus the area under the region

  $y=x^2$

To evaluate ${{\displaystyle \int }}_0^{1.05}\left(x^2+x+1\right)e^{-x}dx-{{\displaystyle \int }}_0^{1.05}{x}^{2}dx$

  

Evaluate ${{\displaystyle \int }}_0^{1.05}\left(x^2+x+1\right)e^{-x}dx$ first, Because it is a product of functions, integration by parts must be used.

  $\begin{array}{c}f(x)=x^2+x+1\\ g(x)=e^{-x}\end{array}$

are calculated using LIPET (Logarithms, Inverse Trig Functions, Polynomials, Exponentials, and Trig Functions). Find the derivatives and anti derivatives, write the plus and minus signs alternately (starting with plus), and then draw arrows connecting each term on the left with the term in the following row on the right to finish the table.

  

The integral of display style ${{\displaystyle \int }}_0^{1.05}\left(x^2+x+1\right)e^{-x}dx$ is written by adding the products of the functions connected by the arrows using the operation sign above each arrow. then condense:

  $\begin{array}{c}{{\displaystyle \int }}_0^{1.05}\left(x^2+x+1\right)e^{-x}dx={\left.\left[\left(x^2+x+1\right)\left(-e^{-x}\right)-(2x+1)\left(e^{-x}\right)+2\left(-e^{-x}\right)\right]\right|}_0^{1.05}\\ =-{\left.e^{-x}\left[x^2+x+1+2(x+1)+2\right]\right|}_0^{1.05}\\ =-{\left.e^{-x}\left(x^2+x+1+2x+2+1\right)\right|}_0^{1.05}\\ =-{\left.e^{-x}\left(x^2+3x+4\right)\right|}_0^{1.05}\\ =-e^{-1.05}\left[{1.05}^2+3(1.05)+4\right]+e^0\left[0^2+3(0)+4\right]\\ \approx -2.888+4\\ =1.112\end{array}$

Now find to use the power rule:

  $\begin{array}{c}{{\displaystyle \int }}_0^{1.05}{x}^{2}dx={\left.\frac{1}{3}{x}^3\right|}_0^{1.05}\\ =\frac{1}{3}\left({1.05}^3-0^3\right)\\ \approx 0.386\end{array}$

To get the area of the region, use the values of each integral:

  $\begin{array}{c}{{\displaystyle \int }}_0^{1.05}\left(x^2+x+1\right)e^{-x}dx-{{\displaystyle \int }}_0^{1.05}{x}^{2}dx\\ \approx 1.112-0.386\\ =0.726\end{array}$

Therefore the required area of the given information is $0.726$ .