All,

I am working on an example problem from "Forgotten Calculus".  This
time on page 352.  The example is the following:

"Find the area between the x-axis and the curve f(x) from x=0 to x=4."

This section of the book is dealing with integrating definite
integrals, so this one shows a function where we'd have to integrate
with the limits of integration being 0 & 4.

When the book graphs the problem, it forms a right triangle.  The
calculated area by the book turns out is shown as 40 square units
(since they are finding area)

The book then poses this question: What would have been the answer if
you had used the formula from geometry for finding the area of a
triangle?

Using A=1/2bh, I come up with 16 square units.

Why the difference in area between the calculation methods???

Thanks,

Herkdrvr

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Clarification of Question by herkdrvr-ga on 14 Mar 2005 14:27 PST

Oops...it should say

"Find the area between the x-axis and the curve f(x)= 5x from x=0 to x=4."

Thanks,

Herkdrvr

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Clarification of Question by herkdrvr-ga on 14 Mar 2005 20:30 PST

One other thing...using 1/2 bh I got 32 units.

The height was depicted as 16 in the book.  Perhaps this is an error
in the book and not on my part--although captaininsano gave me the
first clue.

Thanks...

Herkdrvr

I believe you miscalculated when using A=1/2bh.  The base is equal to
the change in x, which is 4-0=4, so b=4.  The height of the triangle
is f(4)=5(4)=20, so h=20.  Then, A=(1/2)(4)(20)=40 which is the same
as the definite integral.