Riemann Sum Approximations: When are trapezoids more accurate than the middle sum?Which type of Riemann Sum...
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Riemann Sum Approximations: When are trapezoids more accurate than the middle sum?
Which type of Riemann Sum is the most accurate?Similarity between Trapezoidal Sum and Riemann SumsComposite midpoint and trapezoid quadrature of twice differentiable functionRelation between Simpson's Rule, Trapezoid Rule and Midpoint RuleRiemann Integrable Functions SequenceWhy does Trapezoidal Rule have potential error greater than Midpoint?Define the Riemann integral via trapezoids instead of rectanglesFinding $n$ value for trapezoid and midpoint rule errorsWhen calculating riemann sums, which method is more accurate?If the left Riemann sum of a function converges, is the function integrable?Understanding ApproximationSimilarity between Trapezoidal Sum and Riemann Sums
$begingroup$
We can approximate a definite integral, $int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ "tends to be more accurate" then $T_n$?
If $f''$ is continuous on $[a, b]$ and $|f''(x)| leq K$, $forall$ $x in [a, b]$. Then,
$left| int_a^b f(x)dx - T_n right| leq Kfrac{(b-a)^3}{12n^2}$
and
$left| int_a^b f(x)dx - M_n right| leq Kfrac{(b-a)^3}{24n^2}$
For this question let,
$E_{T_n} = left| int_a^b f(x)dx - T_n right| $
and
$E_{M_n} = left| int_a^b f(x)dx - T_n right| $
The theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.
So, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $forall n$?
calculus numerical-methods riemann-sum approximate-integration
edited Mar 19 at 4:20
RRL
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
$endgroup$
add a comment |
$begingroup$
We can approximate a definite integral, $int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ "tends to be more accurate" then $T_n$?
If $f''$ is continuous on $[a, b]$ and $|f''(x)| leq K$, $forall$ $x in [a, b]$. Then,
$left| int_a^b f(x)dx - T_n right| leq Kfrac{(b-a)^3}{12n^2}$
and
$left| int_a^b f(x)dx - M_n right| leq Kfrac{(b-a)^3}{24n^2}$
For this question let,
$E_{T_n} = left| int_a^b f(x)dx - T_n right| $
and
$E_{M_n} = left| int_a^b f(x)dx - T_n right| $
The theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.
So, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $forall n$?
calculus numerical-methods riemann-sum approximate-integration
edited Mar 19 at 4:20
RRL
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
$endgroup$
$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
$begingroup$
This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
$begingroup$
Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56
add a comment |
$begingroup$
We can approximate a definite integral, $int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ "tends to be more accurate" then $T_n$?
If $f''$ is continuous on $[a, b]$ and $|f''(x)| leq K$, $forall$ $x in [a, b]$. Then,
$left| int_a^b f(x)dx - T_n right| leq Kfrac{(b-a)^3}{12n^2}$
and
$left| int_a^b f(x)dx - M_n right| leq Kfrac{(b-a)^3}{24n^2}$
For this question let,
$E_{T_n} = left| int_a^b f(x)dx - T_n right| $
and
$E_{M_n} = left| int_a^b f(x)dx - T_n right| $
The theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.
So, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $forall n$?
calculus numerical-methods riemann-sum approximate-integration
edited Mar 19 at 4:20
RRL
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
$endgroup$
We can approximate a definite integral, $int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ "tends to be more accurate" then $T_n$?
If $f''$ is continuous on $[a, b]$ and $|f''(x)| leq K$, $forall$ $x in [a, b]$. Then,
$left| int_a^b f(x)dx - T_n right| leq Kfrac{(b-a)^3}{12n^2}$
and
$left| int_a^b f(x)dx - M_n right| leq Kfrac{(b-a)^3}{24n^2}$
For this question let,
$E_{T_n} = left| int_a^b f(x)dx - T_n right| $
and
$E_{M_n} = left| int_a^b f(x)dx - T_n right| $
The theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.
So, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $forall n$?
calculus numerical-methods riemann-sum approximate-integration
calculus numerical-methods riemann-sum approximate-integration
edited Mar 19 at 4:20
RRL
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
edited Mar 19 at 4:20
RRL
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
edited Mar 19 at 4:20
RRL
53.4k52574
edited Mar 19 at 4:20
RRL
53.4k52574
edited Mar 19 at 4:20
RRL
53.4k52574
53.4k52574
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
asked May 4 '16 at 7:53
futurebirdfuturebird
3,60412752
3,60412752
$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
$begingroup$
This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
$begingroup$
Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56
add a comment |
$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
$begingroup$
This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
$begingroup$
Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56
$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
$begingroup$
This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
$begingroup$
This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
$begingroup$
Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56
$begingroup$
Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - int_a^b f(x) , dx = int_a^b [f(c) - f(x)] , dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + frac{1}{2} f''(xi_x)(x-c)^2, $$
we see
$$E_M = -int_a^b f'(x)(x-c) , dx + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
Applying integration by parts to the first integral on the RHS we get
$$int_a^b f'(x)(x-c) = left.(x-c)f(x)right|_a^b - int_a^b f(x) , dx = frac{h}{2}[f(a) + f(b)] - int_a^b f(x) , dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = begin{cases} 0.25 + 0.75exp(-200 x^2), &mbox{if } -1 leqslant x leqslant 0 \ 0.99 + 0.01 cos(pi x), &mbox{if } ,,,,,,, 0 < x leqslant 1 end{cases}.$$
Then
$$begin{align}int_{-1}^1 f(x) , dx &approx 1.2870\ M &approx 2 \ T &approx 1.2300 \ E_M &approx 0.7130 \ E_T &approx 0.0570 end{align}$$
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
add a comment |
Your Answer
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$begingroup$
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - int_a^b f(x) , dx = int_a^b [f(c) - f(x)] , dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + frac{1}{2} f''(xi_x)(x-c)^2, $$
we see
$$E_M = -int_a^b f'(x)(x-c) , dx + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
Applying integration by parts to the first integral on the RHS we get
$$int_a^b f'(x)(x-c) = left.(x-c)f(x)right|_a^b - int_a^b f(x) , dx = frac{h}{2}[f(a) + f(b)] - int_a^b f(x) , dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = begin{cases} 0.25 + 0.75exp(-200 x^2), &mbox{if } -1 leqslant x leqslant 0 \ 0.99 + 0.01 cos(pi x), &mbox{if } ,,,,,,, 0 < x leqslant 1 end{cases}.$$
Then
$$begin{align}int_{-1}^1 f(x) , dx &approx 1.2870\ M &approx 2 \ T &approx 1.2300 \ E_M &approx 0.7130 \ E_T &approx 0.0570 end{align}$$
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
add a comment |
$begingroup$
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - int_a^b f(x) , dx = int_a^b [f(c) - f(x)] , dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + frac{1}{2} f''(xi_x)(x-c)^2, $$
we see
$$E_M = -int_a^b f'(x)(x-c) , dx + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
Applying integration by parts to the first integral on the RHS we get
$$int_a^b f'(x)(x-c) = left.(x-c)f(x)right|_a^b - int_a^b f(x) , dx = frac{h}{2}[f(a) + f(b)] - int_a^b f(x) , dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = begin{cases} 0.25 + 0.75exp(-200 x^2), &mbox{if } -1 leqslant x leqslant 0 \ 0.99 + 0.01 cos(pi x), &mbox{if } ,,,,,,, 0 < x leqslant 1 end{cases}.$$
Then
$$begin{align}int_{-1}^1 f(x) , dx &approx 1.2870\ M &approx 2 \ T &approx 1.2300 \ E_M &approx 0.7130 \ E_T &approx 0.0570 end{align}$$
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
$endgroup$
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
add a comment |
$begingroup$
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - int_a^b f(x) , dx = int_a^b [f(c) - f(x)] , dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + frac{1}{2} f''(xi_x)(x-c)^2, $$
we see
$$E_M = -int_a^b f'(x)(x-c) , dx + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
Applying integration by parts to the first integral on the RHS we get
$$int_a^b f'(x)(x-c) = left.(x-c)f(x)right|_a^b - int_a^b f(x) , dx = frac{h}{2}[f(a) + f(b)] - int_a^b f(x) , dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = begin{cases} 0.25 + 0.75exp(-200 x^2), &mbox{if } -1 leqslant x leqslant 0 \ 0.99 + 0.01 cos(pi x), &mbox{if } ,,,,,,, 0 < x leqslant 1 end{cases}.$$
Then
$$begin{align}int_{-1}^1 f(x) , dx &approx 1.2870\ M &approx 2 \ T &approx 1.2300 \ E_M &approx 0.7130 \ E_T &approx 0.0570 end{align}$$
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
$endgroup$
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - int_a^b f(x) , dx = int_a^b [f(c) - f(x)] , dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + frac{1}{2} f''(xi_x)(x-c)^2, $$
we see
$$E_M = -int_a^b f'(x)(x-c) , dx + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
Applying integration by parts to the first integral on the RHS we get
$$int_a^b f'(x)(x-c) = left.(x-c)f(x)right|_a^b - int_a^b f(x) , dx = frac{h}{2}[f(a) + f(b)] - int_a^b f(x) , dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + frac{1}{2}int_a^b f''(xi_x)(x-c)^2 , dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = begin{cases} 0.25 + 0.75exp(-200 x^2), &mbox{if } -1 leqslant x leqslant 0 \ 0.99 + 0.01 cos(pi x), &mbox{if } ,,,,,,, 0 < x leqslant 1 end{cases}.$$
Then
$$begin{align}int_{-1}^1 f(x) , dx &approx 1.2870\ M &approx 2 \ T &approx 1.2300 \ E_M &approx 0.7130 \ E_T &approx 0.0570 end{align}$$
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
edited May 5 '16 at 2:32
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
answered May 5 '16 at 2:18
RRLRRL
53.4k52574
53.4k52574
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
add a comment |
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
$begingroup$
I think you can use periodic smooth functions to get non-contrived examples.
$endgroup$
– Ian
May 5 '16 at 2:49
add a comment |
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$begingroup$
For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $int_0^1 f(x), dx = 0$, but $f(1/2) ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$).
$endgroup$
– Erick Wong
May 4 '16 at 8:40
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This shows $T = 0$, not $E_T = 0$.
$endgroup$
– RRL
May 5 '16 at 2:20
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Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
$endgroup$
– Robert Wolfe
Oct 30 '17 at 15:56