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I preferred this easy calculus train

I preferred this easy calculus train

2023-04-16 20:24:49

I liked this simple calculus exercise

A recent Math SE question
requested for assist computing the worth of
$$int_0^{2000} e^{x/2-leftlfloor x/2rightrfloor}; dx.tag{$star$}$$

Usually after I see somebody’s homework issues I exclaim “what blockhead
TA assigned this?” However I believe it is a actually good train.
Here is why.

In a calculus class, some individuals could have realized to combine frequent
capabilities by rote manipulatation of the expressions. They’ve
realized a algorithm for changing $$int_a^b x^okay; dx$$ to
$$left.frac{x^{okay+1}}{okay+1}rightrvert_a^b$$ after which to
$$frac{b^{okay+1}}{okay+1}- frac{a^{okay+1}}{okay+1}$$ and such like, they usually grind
by the algebra. If that is all somebody is aware of the right way to do, they’re
going to have a variety of bother with !!(star)!!. They could say “However
no person ever taught us the right way to combine capabilities with !!leftlfloor
frac x2rightrfloor!!”.

A calculus tyro making an attempt to take care of this analytically may additionally strive
rewriting $$e^{x/2-leftlfloor x/2rightrfloor}$$ as
$$frac{e^{x/2}}{e^{leftlfloor x/2rightrfloor}}$$ however that makes
the issue more durable, not simpler.

To unravel this, the coed has to truly perceive what the
integral is computing, and if they do not they must be taught
one thing about it. The integral is computing the realm below a curve.
in the event you graph the perform $$frac x2-leftlfloor frac

you discover that it seems to be like this:

The part between 0
and 2 is a straight diagonal
line from ⟨0,0⟩ to ⟨1, 2⟩, and this line repeats over and over on
every interval of length 2.

If the interval of integration in !!(star)!! had been solely !!(0,2)!!
as an alternative of !!(0, 2000)!!, the issue could be
very straightforward as a result of, on this interval, the
sophisticated exponent is identically equal to !!frac

int_0^2 e^{x/2-leftlfloor x/2rightrfloor}; dx
& = int_0^2 e^{x/2}; dx
& = left. 2e^{x/2} rightrvert_0^2
& = 2e-2

Because the perform is totally periodic, integrating over any of
the !!1000!! intervals of size !!2!! will produce the identical worth, so
the ultimate reply is just $$1000cdot (2e-2).$$

However simply pushing across the symbols will not get you there, to resolve this
drawback you must really know one thing about calculus.

The scholar who overcomes this drawback may be taught the next
helpful methods:

  • If some expression seems to be sophisticated, strive graphing it and see in the event you
    get any perception into the way it behaves.

    See Also

  • Some sophisticated capabilities might be understood by breaking them into
    easy elements and coping with the elements individually.

  • Piecewise-continuous capabilities might be built-in by breaking them
    into steady intervals and integrating the intervals individually.

  • You’ll be able to exploit symmetry to cut back the quantity of calculation

None of that is deep stuff, nevertheless it’s all useful method. Additionally
they could make the precious remark that not each drawback ought to
be solved by pushing across the symbols.

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