B1 中級 美國腔 27 分類 收藏
In this video we will review some examples of how to calculate the definite integrals
of types that might show up in an analysis of kinetics. The first is integral of a differential
and I've written this as concentration of B since that's often how it turns out, divided
by 6 times the concentration of B between these two limits. The 6 is a constant, the
integral is just the log and then we're going to evaluate between these two limits. This
is the log of 3 minus the log of 1. This then is the log of 3 minus the log of 1 is 0.
So this definite integral is 0.183. The second integral is the concentration of A to the
- 1/2 power. So the integral becomes plus 1/2 divided by 1/2 and then we have to evaluate
this between the two limits, 1 and 2. So the 1/2 in the denominator 2 and the concentration
of A^1/2 so that's 2^1/2 minus the other limit 1^1/2 and so this definite
integral is 0.83. The third integral, we're going to solve by partial fractions. In order
to do that what we're going to do is take the expression and what we want to do is split
it into two terms because we know how to integrate these two terms. So what we're going to
determine is what are the values of A and B so we can split this into the two terms.
We're going to do that by recombining the right side and compare it to the left side
and get a common denominator for the first term and the second term would be 1 - x.
So what we're going to do is multiply this out. So in order for the right side to look
like the left side, A plus B must equal 1 and 2A plus B is 0 and we're just going to
solve these equations simultaneously. If we do that then A is -1 and B is 2. So we then
can go back and substitute this expression with A is -1 and B is 2 and carry out the integration.
So I have substituted in for A and B to create the two integrals we're
now going to calculate and so in general dx over 1 plus alpha x. We're going to integrate
that, it's the log of 1 plus alpha x divided by alpha. So if we apply this, these two integrals,
we have the log of 1 - x divided by -1 and then we have this minus sign that appears
here and this is evaluated between the limits 0.2 and 0.4. Similarly we have the log of
1 minus 2x divided by -2. And then we also have this 2 that appears here and evaluated
between these limits and if we substitute in and calculate we now have this definite
integral 0.81.


反應動力學-數學準備 (Math Review for Kinetics)

27 分類 收藏
蔡育德 發佈於 2019 年 11 月 9 日
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