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Show using the formal, limited based definition of integral
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I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $lim_{n rightarrow infty}sum_{i=0}^{n-1}f(t_i)Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?
calculus limits definite-integrals
asked Jan 20 at 10:53
UnheilUnheil
83
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add a comment |
$begingroup$
I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $lim_{n rightarrow infty}sum_{i=0}^{n-1}f(t_i)Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?
calculus limits definite-integrals
asked Jan 20 at 10:53
UnheilUnheil
83
$endgroup$
add a comment |
$begingroup$
I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $lim_{n rightarrow infty}sum_{i=0}^{n-1}f(t_i)Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?
calculus limits definite-integrals
asked Jan 20 at 10:53
UnheilUnheil
83
$endgroup$
I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $mathbb{R}$). However, I already seem to get stuck on the formal definition of the integral, which is (I believe) $lim_{n rightarrow infty}sum_{i=0}^{n-1}f(t_i)Delta t$. Can anyone explain to me how to proceed? Or show me the steps I need to follow to get to the right answer?
calculus limits definite-integrals
calculus limits definite-integrals
asked Jan 20 at 10:53
UnheilUnheil
83
asked Jan 20 at 10:53
UnheilUnheil
83
edited Jan 20 at 11:41
Unheil
asked Jan 20 at 10:53
UnheilUnheil
83
asked Jan 20 at 10:53
UnheilUnheil
83
asked Jan 20 at 10:53
UnheilUnheil
83
83
add a comment |
add a comment |
1 Answer
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Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= frac{4}{n}+ frac{32i}{n^2}$. Add those: $sum_{i= 0}^{n-1} frac{4}{n}+ frac{32i}{n^2}= frac{1}{n}sum_{i=0}^{n-1} 4+ frac{32}{n^2}sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $frac{32}{n^2}$: $frac{32}{n^2}frac{n(n-1)}{2}= frac{32}{n^2}frac{n^2- n}{2}= 16(1- frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
answered Jan 20 at 12:29
user247327user247327
11.2k1515
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Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
add a comment |
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$begingroup$
Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= frac{4}{n}+ frac{32i}{n^2}$. Add those: $sum_{i= 0}^{n-1} frac{4}{n}+ frac{32i}{n^2}= frac{1}{n}sum_{i=0}^{n-1} 4+ frac{32}{n^2}sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $frac{32}{n^2}$: $frac{32}{n^2}frac{n(n-1)}{2}= frac{32}{n^2}frac{n^2- n}{2}= 16(1- frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
answered Jan 20 at 12:29
user247327user247327
11.2k1515
$endgroup$
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
add a comment |
$begingroup$
Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= frac{4}{n}+ frac{32i}{n^2}$. Add those: $sum_{i= 0}^{n-1} frac{4}{n}+ frac{32i}{n^2}= frac{1}{n}sum_{i=0}^{n-1} 4+ frac{32}{n^2}sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $frac{32}{n^2}$: $frac{32}{n^2}frac{n(n-1)}{2}= frac{32}{n^2}frac{n^2- n}{2}= 16(1- frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
answered Jan 20 at 12:29
user247327user247327
11.2k1515
$endgroup$
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
add a comment |
$begingroup$
Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= frac{4}{n}+ frac{32i}{n^2}$. Add those: $sum_{i= 0}^{n-1} frac{4}{n}+ frac{32i}{n^2}= frac{1}{n}sum_{i=0}^{n-1} 4+ frac{32}{n^2}sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $frac{32}{n^2}$: $frac{32}{n^2}frac{n(n-1)}{2}= frac{32}{n^2}frac{n^2- n}{2}= 16(1- frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
answered Jan 20 at 12:29
user247327user247327
11.2k1515
$endgroup$
Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)/n- 5= (6n+ 8i- 5n)/n= (n+ 8i)/n$.
On the ith interval, $f(x_i)Delta x= [(n+ 8i)/n](4/n)= (4n+ 32i)/n^2= frac{4}{n}+ frac{32i}{n^2}$. Add those: $sum_{i= 0}^{n-1} frac{4}{n}+ frac{32i}{n^2}= frac{1}{n}sum_{i=0}^{n-1} 4+ frac{32}{n^2}sum_{i=0}^{n-1} i$.
Now, adding "4" to itself n times is just 4n while adding "i", with i running from 0 to n-1 means 0+ 1+ 2+ 3+ 4+ …+ n-1.
So what is such sum? To start with lets call it S(n). That is, S(n)= 0+ 1+ 2+ 3+ 4+ …+ n-1. Of course, it is the same if we do the sum in the opposite order: S(n)= n-1+ n-2+ n-3+ n-4+...+ 1+ 0. Add those two "vertically"! That is, 2S(n)= [0+ n-1]+ [1+ n-2]+ [2+ n-3]+ …+ [n-2+ 1]+ [n- 1]. Every term is "n-1" and there are n terms. 2S= n(n-1) so S= n(n-1)/2.
Check that for a few values of n: with n= 5, 0+ 1+ 2+ 3+ 4= 10 and 5(5-1)/2= 20/2= 10. with n= 9, 0+ 1+ 2+ 3+ 4+ 5+ 6+ 7+ 8= 36 and 9(8)/2= 36.
So we have that the sum is n(n-1)/2 and we multiply that by $frac{32}{n^2}$: $frac{32}{n^2}frac{n(n-1)}{2}= frac{32}{n^2}frac{n^2- n}{2}= 16(1- frac{1}{n})$.
Putting those together the "Riemann sum" with n intervals is $4+ 16(1- frac{1}{n})$. As n goes to infinity, 1/n goes to 0 so that sum goes to $4+ 16(1+ 0)= 20$.
Of course, geometrically, the area under that straight line from x= 3 to x= 7 is a trapezoid with bases of length 2(3)- 5= 1 and 2(7)- 5= 9 and height 7- 3= 4. The area of that trapezoid is (1/2)(1+ 9)(4)= 20 as before.
answered Jan 20 at 12:29
user247327user247327
11.2k1515
answered Jan 20 at 12:29
user247327user247327
11.2k1515
answered Jan 20 at 12:29
user247327user247327
11.2k1515
answered Jan 20 at 12:29
user247327user247327
11.2k1515
11.2k1515
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
add a comment |
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
$begingroup$
Thank you very much, I completely understand it now!
$endgroup$
– Unheil
Jan 20 at 13:29
add a comment |
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