Integration Problem

James10492

New member
Joined
May 17, 2020
Messages
8
Hi there, I am stuck on this problem and would very grateful if someone could help me out.

10 (axa) dx = a/a+1

I don't know how this solution has been arrived at. When I try myself I get aa+1/a+1 (because a times 0 is zero, and anything divided by zero is zero, so the value of the lower bound is neutral).

This is how I have solved all definite integration problems up to now. What am I doing wrong?
 

lex

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Joined
Mar 3, 2021
Messages
480
\(\displaystyle \int_{x=0}^{x=1}{{ax}^a dx}\)

\(\displaystyle =a\int_{x=0}^{x=1}{x^a dx}\)

\(\displaystyle =a \left[\frac{x^{a+1}}{a+1}\right]^{x=1}_{x=0} \quad \text{with } a>-1\)

\(\displaystyle =\frac{a}{a+1}\left(\left(1^{a+1}-0^{a+1}\right)\right)\)

\(\displaystyle =\frac{a}{a+1}, \quad a>-1\)

1. You seem to have been substituting \(\displaystyle a\) in for \(\displaystyle x\). That's why I explicitly put in \(\displaystyle x=1\) and \(\displaystyle x=0\).
2. Re. your statement: "and anything divided by zero is zero, so the value of the lower bound is neutral". The first part is not true. You cannot divide by 0. (There is no answer, 0 or otherwise). The second half of the statement I don't understand.
3. As you see from my answer, this integral only exists for \(\displaystyle a>-1\). To get an idea of what's happening you could graph \(\displaystyle x^a\) for some negative values of \(\displaystyle a\). (Even though, this doesn't tell the whole story).
 

James10492

New member
Joined
May 17, 2020
Messages
8
Thank you, I think I understand now.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
10,405
anything divided by zero is zero.

Let's see if that is true.

10/2 =5 since only 2*5 = 10
24/4 = 6 since only 4*6 = 24

7/0 = 0 since only 0*0 = 7

12/0 = 0 since only 0*0 = 12.

You are correct IF you believe that 0*0=7 and 0*0 = 12
 
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