Fundamental Theorem of Calculus Part II

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Hckyplayer8

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I left the two terms separate cause I figured the Algebra needed to combine the two wasn't worth it considering my upper and lower bounds were [-1,1]. Therefore (pending my antiderivative is correct) I can plug in my bounds which will eliminate the powers since 1 to any power will be 1. Then it is just a matter of summing the fractions.
 
Your third line doesn't make much sense as written. (Is that something you were taught to write, or just your own self-invented scratch work?) But you evidently meant the right thing, because the final line is correct. And you're right not to bother combining terms, which would be a waste of effort.

So, finish it up!
 
Hckyplayer8 said:
View attachment 14618

I left the two terms separate cause I figured the Algebra needed to combine the two wasn't worth it considering my upper and lower bounds were [-1,1]. Therefore (pending my antiderivative is correct) I can plug in my bounds which will eliminate the powers since 1 to any power will be 1. Then it is just a matter of summing the fractions.
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I thought that we agreed that it is a waste of time to write the C IF it is a definite integral.
 
Here is a nice trick to easily add 1 to a fraction.

\(\displaystyle \frac {a}{b} + 1 = \frac {a}{b} + \frac {b}{b} = \frac {a+b}{b}\). That is the numerator of the sum is simply the sum of the numerator and denominator that you are adding 1 to, divided by the same denominator
 
Jomo said:
I thought that we agreed that it is a waste of time to write the C IF it is a definite integral.
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It is actually incorrect to write \(\displaystyle +C\) in an answer to a definite integral.
A definite integral is a number. There is no ambiguity about it.
 
pka said:
It is actually incorrect to write \(\displaystyle +C\) in an answer to a definite integral.
A definite integral is a number. There is no ambiguity about it.
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That was exactly why I told hckeyplayer8 that it was a waste of time to write +C and -C as they will cancel out in every definite integral.
 
Dr.Peterson said:
Your third line doesn't make much sense as written. (Is that something you were taught to write, or just your own self-invented scratch work?) But you evidently meant the right thing, because the final line is correct. And you're right not to bother combining terms, which would be a waste of effort.

So, finish it up!
Click to expand...

The third line was more or less in effort to show my work than anything.
 
Jomo said:
Here is a nice trick to easily add 1 to a fraction.

\(\displaystyle \frac {a}{b} + 1 = \frac {a}{b} + \frac {b}{b} = \frac {a+b}{b}\). That is the numerator of the sum is simply the sum of the numerator and denominator that you are adding 1 to, divided by the same denominator
Click to expand...

These quick Algebraic tricks are definitely a weak spot in my math game. Appreciate the post.
 
Dr.Peterson said:
You misplaced some parentheses. What is [MATH](-1)^{4/3}[/MATH]?
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If it's not -1 then it has to be 1. I blanked on bases raised to an improper fraction so I had to look it up. Afterwards I see that term is just the cube root of negative one raised to the fourth power which results in positive one.

Thank you for pointing that out.

What do you mean by misplaced parentheses?
 
Hckyplayer8 said:
If it's not -1 then it has to be 1. I blanked on bases raised to an improper fraction so I had to look it up. Afterwards I see that term is just the cube root of negative one raised to the fourth power which results in positive one.

Thank you for pointing that out.

What do you mean by misplaced parentheses?
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\(\displaystyle -1^\frac{4}{3} \neq (-1)^\frac{4}{3}\). Just like \(\displaystyle -4 = -2^2 \neq (-2)^2 = 4\), so you needed the parenthesis around the -1 !

In computing \(\displaystyle (-1)^\frac{4}{3}\) you have two options. You can compute \(\displaystyle (-1)^4\), getting 1 and then compute \(\displaystyle \sqrt[3]{1} = 1\)

Or you can compute \(\displaystyle \sqrt[3]{-1} = -1\), then compute \(\displaystyle (-1)^4 = 1\)
 
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