Indefinite Integrals Without Differentation and Fundamental Theorem
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As you know we can solve definite integrals using the definition over limit (without using antiderivatives at all). So I what I ask is how to solve an indefinite integral like that? For example $$int x^2 , dx$$ , we all know the result is x^3/3 + c. But we got this by using differentation and Fundamental Theorem of Calculus. How can we calculate this without using them?
calculus integration
asked Jan 6 at 16:15
madscientistmadscientist
11
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add a comment |
$begingroup$
As you know we can solve definite integrals using the definition over limit (without using antiderivatives at all). So I what I ask is how to solve an indefinite integral like that? For example $$int x^2 , dx$$ , we all know the result is x^3/3 + c. But we got this by using differentation and Fundamental Theorem of Calculus. How can we calculate this without using them?
calculus integration
asked Jan 6 at 16:15
madscientistmadscientist
11
$endgroup$
$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43
add a comment |
$begingroup$
As you know we can solve definite integrals using the definition over limit (without using antiderivatives at all). So I what I ask is how to solve an indefinite integral like that? For example $$int x^2 , dx$$ , we all know the result is x^3/3 + c. But we got this by using differentation and Fundamental Theorem of Calculus. How can we calculate this without using them?
calculus integration
asked Jan 6 at 16:15
madscientistmadscientist
11
$endgroup$
As you know we can solve definite integrals using the definition over limit (without using antiderivatives at all). So I what I ask is how to solve an indefinite integral like that? For example $$int x^2 , dx$$ , we all know the result is x^3/3 + c. But we got this by using differentation and Fundamental Theorem of Calculus. How can we calculate this without using them?
calculus integration
calculus integration
asked Jan 6 at 16:15
madscientistmadscientist
11
asked Jan 6 at 16:15
madscientistmadscientist
11
asked Jan 6 at 16:15
madscientistmadscientist
11
asked Jan 6 at 16:15
madscientistmadscientist
11
asked Jan 6 at 16:15
madscientistmadscientist
11
11
$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43
add a comment |
$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43
$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43
$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Divide the interval into n parts and use
$sum_{k=1}^n k^2 =n(n+1)(2n+1)/6
=n^3/3+$ smaller terms.
Take the limit.
This is how these integrals were originally found.
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
$endgroup$
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Divide the interval into n parts and use
$sum_{k=1}^n k^2 =n(n+1)(2n+1)/6
=n^3/3+$ smaller terms.
Take the limit.
This is how these integrals were originally found.
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
$endgroup$
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
add a comment |
$begingroup$
Divide the interval into n parts and use
$sum_{k=1}^n k^2 =n(n+1)(2n+1)/6
=n^3/3+$ smaller terms.
Take the limit.
This is how these integrals were originally found.
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
$endgroup$
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
add a comment |
$begingroup$
Divide the interval into n parts and use
$sum_{k=1}^n k^2 =n(n+1)(2n+1)/6
=n^3/3+$ smaller terms.
Take the limit.
This is how these integrals were originally found.
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
$endgroup$
Divide the interval into n parts and use
$sum_{k=1}^n k^2 =n(n+1)(2n+1)/6
=n^3/3+$ smaller terms.
Take the limit.
This is how these integrals were originally found.
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
answered Jan 6 at 16:24
marty cohenmarty cohen
75.1k549130
75.1k549130
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
add a comment |
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
$begingroup$
Can you write the exact limit expression? It seems easy this way but I need it to understand for real.
$endgroup$
– madscientist
Jan 7 at 18:15
add a comment |
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$begingroup$
Indefinite integrals are by definition anti-derivatives. There is no direct limit definition for them.
$endgroup$
– Paramanand Singh
Jan 6 at 16:43