Name: Math Review for Kinetics
Uploaded: 2019-11-09T05:54:48.000Z
Duration: 3 min 46 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

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

Modal verbs - B1

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