# Intro-Calculus integral problem anyone?

1. Dec 26, 2006

### rainyrabbit

1. The problem statement, all variables and given/known data

Finding Area Find the area of the region enclosed by the y-axis and the curves y=x^2 and y=(x^2+x+1)*e^(-x)

The problem is from intro-level calculus and it's from
Chapter 6: Differential Equations and Mathematical Modeling
Section 3: Integration by parts
Book: Calculus: graphical, numerical, algebraic. By Finney, Demmana, et. al

Please, help. I kinda do not understand the problem.

2. Relevant equations

3. The attempt at a solution

Int(y)dy = Int(x^2)dy.... Did not go further.
I don't really know how to do it.

2. Dec 26, 2006

### Gregie666

http://img186.imageshack.us/img186/2536/graphuq4.th.jpg [Broken]
take a look at the picture. you need to find the area enclosed between the two curves and the y axis. lets call this area S.
first find the x value of the point of intersection of your two functions. lets call it b.
then find the area under the upper curve in the stretch of the x axis that goes from 0 to b. lets call this S1.
then find the area under the bottom curve in the same stretch of the x area
[0, b]. lets call this S2. now, its clear that S1-S2=S which is the area you need to find.
remeber that the area under a function in the closed stretch of the x axis that goes from a to b aka [a,b] is:
$$\int_a^b {f(x)dx}$$
!!!there are exceptions to this rule(but not in this case)!!!

Last edited by a moderator: May 2, 2017
3. Dec 26, 2006

### HallsofIvy

Staff Emeritus
On its face, its a very straight forward problem: draw a graph, determine exactly what region you are talking about. Find, either horizontally or vertically, straight lines that cover the region. The integral will be the length of those straight lines times either dx or dy.

2. Relevant equations

3. The attempt at a solution

Int(y)dy = Int(x^2)dy.... Did not go further.
I don't really know how to do it.[/QUOTE]
And what you did is wrong- you've ignored part of the boundary. Obviously y= x2 crosses the y-axis at (0,0). When x= 0 (the y-axis) the other function is e0(1)= 1: the other graph crosses the y-axis at (0, 1). The hard part may be determining where the two graphs intersect, where x2= e-x(x2+ x+ 1). Since your text is titled "Calculus: graphical, numerical, algebraic" you may be expected to do that numerically. Looks like you will have the integral from 0 to whatever the x value is where the two graph intersect of e-x(x2+ x+ 1)- x2.

4. Jan 2, 2007

### rainyrabbit

Thank you for every help you have given to me. I understand now.