Consider the Integral $ int_{0}^1left( x^3-3x^2right)dx $ and evaluate using Riemann Sum
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Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$
$x_i=0+frac{1}{n}i$
Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$
Continuing the problem I get
$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$
which leads me to
$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$
After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$
The answer is $-frac{3}{4}$ but I keep getting 1.
calculus riemann-sum
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
add a comment |
$begingroup$
Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$
$x_i=0+frac{1}{n}i$
Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$
Continuing the problem I get
$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$
which leads me to
$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$
After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$
The answer is $-frac{3}{4}$ but I keep getting 1.
calculus riemann-sum
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
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What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48
add a comment |
$begingroup$
Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$
$x_i=0+frac{1}{n}i$
Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$
Continuing the problem I get
$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$
which leads me to
$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$
After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$
The answer is $-frac{3}{4}$ but I keep getting 1.
calculus riemann-sum
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
$endgroup$
Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$
$x_i=0+frac{1}{n}i$
Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$
Continuing the problem I get
$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$
which leads me to
$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$
After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$
The answer is $-frac{3}{4}$ but I keep getting 1.
calculus riemann-sum
calculus riemann-sum
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
edited Dec 10 '18 at 0:02
Eric Brown
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
asked Dec 9 '18 at 23:16
Eric BrownEric Brown
737
737
$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48
add a comment |
$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48
$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48
$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48
add a comment |
1 Answer
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$begingroup$
I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
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1 Answer
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1 Answer
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$begingroup$
I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
$endgroup$
add a comment |
$begingroup$
I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
$endgroup$
add a comment |
$begingroup$
I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
$endgroup$
I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
answered Dec 10 '18 at 0:05
YiFanYiFan
2,8291422
2,8291422
add a comment |
add a comment |
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$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26
$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48