Finding lower and upper sums for
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$
Find a lower sum for
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval
$\left[a,b\right]=\left[0,\frac{\pi}{2}\right];$ let
$n=6.$
Let’s first look at the graph in
[link] to get a better idea of the area of interest.
The intervals are
$\left[0,\frac{\pi}{12}\right],\left[\frac{\pi}{12},\frac{\pi}{6}\right],\left[\frac{\pi}{6},\frac{\pi}{4}\right],\left[\frac{\pi}{4},\frac{\pi}{3}\right],\left[\frac{\pi}{3},\frac{5\pi}{12}\right],$ and
$\left[\frac{5\pi}{12},\frac{\pi}{2}\right].$ Note that
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ is increasing on the interval
$\left[0,\frac{\pi}{2}\right],$ so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum
$\sum _{i=0}^{5}\text{sin}\phantom{\rule{0.1em}{0ex}}{x}_{i}\left(\frac{\pi}{12}\right)}.$ We have
Using the function
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval
$\left[0,\frac{\pi}{2}\right],$ find an upper sum; let
$n=6.$
The use of sigma (summation) notation of the form
$\sum _{i=1}^{n}{a}_{i}$ is useful for expressing long sums of values in compact form.
For a continuous function defined over an interval
$\left[a,b\right],$ the process of dividing the interval into
n equal parts, extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing the areas yields an approximation of the area of that region.
The width of each rectangle is
$\text{\Delta}x=\frac{b-a}{n}.$
Riemann sums are expressions of the form
$\sum _{i=1}^{n}f\left({x}_{i}^{*}\right)\text{\Delta}x},$ and can be used to estimate the area under the curve
$y=f\left(x\right).$ Left- and right-endpoint approximations are special kinds of Riemann sums where the values of
$\left\{{x}_{i}^{*}\right\}$ are chosen to be the left or right endpoints of the subintervals, respectively.
Riemann sums allow for much flexibility in choosing the set of points
$\left\{{x}_{i}^{*}\right\}$ at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum.
State whether the given sums are equal or unequal.
$\sum _{i=1}^{10}i$ and
$\sum _{k=1}^{10}k$
$\sum _{i=1}^{10}i$ and
$\sum _{i=6}^{15}\left(i-5\right)$
$\sum _{i=1}^{10}i\left(i-1\right)$ and
$\sum _{j=0}^{9}\left(j+1\right)j$
$\sum _{i=1}^{10}i\left(i-1\right)$ and
$\sum _{k=1}^{10}\left({k}^{2}-k\right)$
a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting
$j=i-1.$ d. They are equal; the first sum factors the terms of the second.
for MSc chemistry... simple formulas of integration
aparna
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funny
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funny
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aparna
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aparna
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funny
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funny
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funny
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aparna
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aparna
value of log ax
cot-x cos-x
aparna
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funny
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funny
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aparna
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funny
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aparna
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funny
integration f(X) dx
this is basic formula of integration sign is not there you can look integration sign in methematics form...
and f(X) my be any function any values
funny
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divide by tan^2 x
giving
1=csc^2 x -1/tan^2 x,
rewrite as:
1=1/sin^2 x -cos^2 x/sin^2 x,
multiply by sin^2 x giving:
sin^2 x=1-cos^2x.
rewrite as the familiar
sin^2 x + cos^2x=1
QED
sin x - sin x cos^2 x
sin x (1-cos^2 x)
note the identity:sin^2 x + cos^2 x = 1
thus, sin^2 x = 1 - cos^2 x
now substitute this into the above:
sin x (sin^2 x), now multiply, yielding:
sin^3 x
Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x
=sinx-sinx(1+sin^2x)
=sinx-sinx+sin^3x
=sin^3x
thats proved.