[maths] Re: integrals; calculus
- From: "Nelson Blachman" <blachman@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Thu, 14 Dec 2006 01:19:57 -0800
Ned,
I get 42 for the integral of (1+3x)dx from -1 to 5 by using
"antiderivatives." What result do you get via the defining limit and how do
you get it?
For integral from 1 to 64 (1+cube_root x) / sqrt x dx
= integral of [x^(-1/2) + x^(-1/6)] dx =
= [2x^(1/2) + (6/5)x^(6/5)] for x=64 minus for x=1= 14 + 37.2 = 51.2
What do you get, and how?
I haven't given any thought yet to how to use the defining limit.
--Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Wednesday, December 13, 2006 11:23 AM
Subject: [maths] integrals; calculus
Hi all,
A couple of very quick questions:
1. Compute the following using the equation as limits of sums from the
definition of the integral.
That equation is:
integral from a to b of f(x) dx
= limit as n approaches infinity of summation (Sigma) from i=1 to n of f(x
sub i) delta x.
The problem itself is:
integral from -1 to 5 of (1+3x) dx.
When I use the FTC2, I get the correct answer 42, but not this way.
2. Use FTC2 to evaluate:
integral from 1 to 64 (1+cube_root x) / sqrt x dx.
I get 32, but the right answer is 256/5.
Many thanks in advance!
Ned
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