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  • 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.

In this video we will review some examples of how to calculate the definite integrals

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B1 中級

運動学のための数学の復習 (Math Review for Kinetics)

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    蔡育德 に公開 2021 年 01 月 14 日
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