## Monday, May 8, 2017

### Unizor - Definite Integrals - Other Examples

Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Other Examples

Distance

Consider the following problem.
A car moves along a straight line with variable speed given by function v(t) that defines speed v at any moment of time t.
Our task is to find the distance covered by this car from time moment t=a to moment t=b.

If the speed is constant v(t)=V, the solution is easy:
S = V·(b−a)

For variable speed the problem is not that easy. Here is what can be suggested as an approximate solution.
Let's divide our time interval [a,b] into N equal short intervals [a=t0,t1], [t1,t2], [t2,t3]... [tN−1,tN=b] and assume that during each interval the speed is not significantly changing - a reasonable assumption if the time interval is small enough.
Then the distance covered during i-th time interval [ti−1,ti] is approximately equal to
ΔSi = v(ti)·(ti−ti−1)
Here we use v(ti) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔSi to get a total distance and go to a limit as the number of intervals we divide our total travel time increases to infinity.
So, the final formula is
S = lim Σi∈[1,N] v(ti)·(ti−ti−1) =
lim Σi∈[1,N] v(tiΔti
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δti diminishes to zero.

As with a problem of area under curve, we can prove that the limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time travel into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total travel time into more and more intervals.

Draining

Consider the following problem.
There is a tub filled with water. When we open the drain, the water starts flowing out of the tub. The speed of water flow depends on different factors - some constant (the size of a drain pipe) and some variable (the pressure at the drain opening).

Our task is to determine the volume of water that is drained from the tub during some known period of time from t=a to t=b, provided we know the speed of draining v(t) (in some units, like liters per second) at any moment of time t.

The complication here, obviously, is that the speed of water flow through a drain is variable because it depends on the water pressure at the drain opening, which changes as the water flows out of the tub.

Our approach to this problem is similar to the one above.
Let's divide our time interval [a,b] into N equal short intervals [a=t0,t1], [t1,t2], [t2,t3]... [tN−1,tN=b] and assume that during each interval the speed of water flow through a drain is not significantly changing - a reasonable assumption if the time interval is small enough.
Then the volume of water going through a drain during i-th time interval [ti−1,ti] is approximately equal to
ΔWi = v(ti)·(ti−ti−1)
Here we use v(ti) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔWi to get a total volume of drained water and go to a limit as the number of intervals we divide our total drainage time increases to infinity.
So, the final formula is
W = lim Σi∈[1,N] v(ti)·(ti−ti−1) =
lim Σi∈[1,N] v(tiΔti
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δti diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time of drainage into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total time into more and more intervals.

Volume of Solids of Revolution

Consider the following problem.
There is a solid obtained by a revolution of some curve on a plane around an axis which also lies on this plane.
Let's assume that the curve is defined as a graph of a smooth function y=f(x), where argument x varies from a to b, and the axis of rotation is the X-axis.

Our task is to determine the volume of this solid.

If our function is constant, that is f(x)=c, our solid is a cylinder of height H=b−a and radius equal to that constant c.
If function f(x) is linear (so, its graph is a straight line), our solid is a truncated cone of height H=b−a and radiuses of its two bases equal to Ra=f(a) and Rb=f(b).
In either of the above cases we know classic geometric formulas to calculate the volume of a solid.
The complexity of our problem, obviously, is that the shape of our solid is not one of those well known types.

Let's approach our problem analogously to determining the area under curve.
Let's divide our argument interval [a,b] into N equal short intervals [a=x0,x1], [x1,x2], [x2,x3]... [xN−1,xN=b] and assume that on each interval the value of function f(x) is not significantly changing - a reasonable assumption if the interval is small enough.

Replacing the function values on each interval [xi−1,xi] with a constant value at its right margin f(xi) and rotating the obtained step-function around the X-axis, we obtain a different solid, but the one that approximates the original one to certain degree.
The approximation will be better if the number of intervals we divide segment [a,b] is increasing and the size of the largest interval diminishes to zero.

The result of rotation of a step-function on each interval will be a cylinder with height hi=xi−xi−1 and radius f(xi), its volume will be equal to
ΔVi = π·f 2(xi)·(xi−xi−1) =
= π·f 2(xi
Δxi

Our next step is to summarize all ΔVi to get a total volume of a solid and go to a limit as the number of intervals we divide our segment [a,b].
So, the final formula is
V = lim Σi∈[1,N] π·f 2(xiΔxi
where limit is assumed to be taken when N→∞ and maximum width among all intervals Δxi diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the radius of a cylinder. This limit is also independent on how we break the total range of arguments into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total range into more and more intervals.

CONCLUSION

In all the cases we considered so far we came up with an expression
lim Σi∈[1,N] f(xiΔxi
where f(x) is some smooth function defined on a segment [a,b],
{xi} is partitioning of segment [a,b] into N parts,
and we assume that N→∞ and the maximum width of intervals Δxi=xi−xi−1 converges to zero.

We have also proven that this limit exists for any smooth function f(x), that it does not depend on how we partition our segment [a,b] (as long as the widest interval of division shrinks to zero length) and it does not depend on which point within each interval of division we use to determine the function value on this interval.