The Area Under a Parabola and the Length of a Curve

Date: 11/15/95 at 21:57:28
From: Navigator
Subject: area under parabola

The following problems are an enigma to me.

Find the area under the parabola f(x) = x^2 + 1 for x = 0 to x = 2, 
and y > 0.
Also find the length of the curve from x = 0 to x = 2.

Find the area under the parabola f(x) = x^2 - 2x + 4 and the horizontal 
axis for x = 0 to x = 3, and y > 0.  Also, find the length of the curve 
from x = 0 to x = 3.

I need help on these problems please. 

John 102344.1331@compuserve.com

Date: 11/19/95 at 17:15:20
From: Doctor Josh
Subject: Re: area under parabola

Dear John,

The area under the first parabola is the integral from 0 to 2 of x^2+1.
Thus, 
           /
    Area = |[the integral sign] (x^2 + 1)dx from 0 to 2
           /
         = [(1/3)(2^3) + 2] - [(1/3)(0^3) + 0]
 
         = 14/3 square units

The length of the curve from x=0 to x=2 is:
               /
      length = |[1 - (y')^2]dx from x=0 to x=2
               /

Draw out a triangle of dx, dy, and dl to see why.

Use the same techniques for the other parabola.

-Doctor Josh,  The Geometry Forum