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If [tex]f[/tex] is a positive function and [tex]f^{\prime \prime}(x) < 0[/tex] for [tex]a\leq x\leq b[/tex], show that
[tex]T_n < \int _a ^b f(x)\: dx < M_n[/tex]
where:
[tex]T_n[/tex] is the Trapezoidal Rule.
[tex]M_n[/tex] is the Midpoint Rule.
In my textbook (Calculus: concepts and contexts / James Stewart. --- 2nd ed. --- page 419), there is a figure with the following caption:
"The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid [tex]ABCD[/tex] whose upper side is tangent to the graph at [tex]P[/tex]. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid [tex]AQRD[/tex] used in the Trapezoidal Rule."
In other words, provided that the concavity remains the same, we can trap the exact value of the integral between the trapezoidal and midpoint sums for any number of subdivisions.
Maybe, it can be used as the answer. However, I think I'd be better to express that in mathematical terms. So, here's what I've got:
which can be written as
Am I on the right track? What should I do next?
Any help is highly appreciated.
[tex]T_n < \int _a ^b f(x)\: dx < M_n[/tex]
where:
[tex]T_n[/tex] is the Trapezoidal Rule.
[tex]M_n[/tex] is the Midpoint Rule.
In my textbook (Calculus: concepts and contexts / James Stewart. --- 2nd ed. --- page 419), there is a figure with the following caption:
"The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid [tex]ABCD[/tex] whose upper side is tangent to the graph at [tex]P[/tex]. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid [tex]AQRD[/tex] used in the Trapezoidal Rule."
In other words, provided that the concavity remains the same, we can trap the exact value of the integral between the trapezoidal and midpoint sums for any number of subdivisions.
Maybe, it can be used as the answer. However, I think I'd be better to express that in mathematical terms. So, here's what I've got:
[tex]\frac{h}{2}\left\{ f(a) + 2f(a+h) + 2f(a+2h) + \dots + 2f\left[ a + (n-2)h \right] + 2f\left[ a + (n-1)h \right] + f\left( a + nh \right) \right\}[/tex]
[tex]<[/tex]
[tex]\int _a ^b f(x)\: dx[/tex]
[tex]<[/tex]
[tex]h \left\{ f\left( a + \frac{h}{2} \right) + f\left( a + \frac{3h}{2} \right) + f\left( a + \frac{5h}{2} \right) + \dots + f\left[ a + \frac{(2n-3)h}{2} \right] + f\left[ a + \frac{(2n-2)h}{2} \right] + f\left[ a + \frac{(2n-1)h}{2} \right] \right\}[/tex]
which can be written as
[tex]\frac{1}{2} f(a) + f(a+h) + f(a+2h) + \dots + f\left[ a + (n-2)h \right] + f\left[ a + (n-1)h \right] + \frac{1}{2} f\left( a + nh \right) [/tex]
[tex]<[/tex]
[tex]\int _a ^b f(x)\: dx[/tex]
[tex]<[/tex]
[tex]f\left( a + \frac{h}{2} \right) + f\left( a + \frac{3h}{2} \right) + f\left( a + \frac{5h}{2} \right) + \dots + f\left[ a + \frac{(2n-3)h}{2} \right] + f\left[ a + \frac{(2n-2)h}{2} \right] + f\left[ a + \frac{(2n-1)h}{2} \right][/tex]
Am I on the right track? What should I do next?
Any help is highly appreciated.
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