Practical lesson 1. Calculating a certain integral
The topic of our lesson is “Calculating a certain integral.” Let’s remember what methods exist for this. Actually, they are the same as for the indefinite integral, but there are some slight differences, of course.
The easiest way is to integrate directly. At this stage, the most important is that we use the formula of Newton-Leibniz. Most often, we write the right part as the difference F(b)-F(a).
However, it will be very convenient for us to use a vertical number to calculate the integral and record it. We say that we will perform the following calculation: F(x), vertical bar, a and b – indexes, lower and upper ones (see the video).
In order to calculate productively and rationally, it is very important to know the properties of such a calculation. In general, it satisfies the linearity property. If such a trait refers to the sum of a function, we can split this sum into terms and perform calculations for each term.
And the second property is the uniformity property. We can take a constant multiplier for the sign of this calculation. This is direct integration and the Newton-Leibniz formula. Other important methods are the same as for an indefinite integral: the method of substitution or replacement of variables and the method of integration by parts. The formula is almost the same, only the integrals have integration limits and the product of the function UV also with a vertical line from a to b.
Let’s look at examples of all these methods.
The first example is very simple, of course (see the video). So, have a look at how to correctly, or rather, rationally perform the calculation of this integral. Calculating the integral of the first term, we get: one third x^3, then minus one second x^2 and plus 4x. According to the Newton-Leibniz formula, we should have written: we perform a calculation from -1 to 2. That is, generally speaking, you can calculate the value of this function at point 2, then write a minus and then calculate the value of the function at point -1. But you will face the fact that at each stage you will have to add and subtract fractions. It is more rational to use the calculation properties.
So, one third is multiplied by x^3 from -1 to 2, then minus one second is multiplied by x^2 from -1 to 2, and then 4 is multiplied by x from -1 to 2. So, let’s see. So, what do we multiply one-third by? Substituting the upper value, we get 2^3 is 8. And subtract how many values at point -1? Minus minus 1 means 8+1. Then, minus one second, at the top point we get 4, at the bottom point we get 1 because (-1)^2, plus 4 multiplied by the difference between the values of 2 and –(-1), 2+1, i.e. by 3. See what happens. Here the fractions have disappeared altogether, this is 3. Here 3 divided by 2 is 1.5. And here +12. So, we get 13.5 (see the video).
So, let’s look at the following example. We analyze the situation immediately. Under the sign of the integral is the product of two functions that are not related to each other by differentiation, that is, we can’t put anything under the sign of the differential. And what remains for us, apparently, in this case, to apply the formula of integration in parts. Let us remember: UdV is UV minus the integral of VdU. This is if there was an indefinite integral, we would stop there. But we have a definite integral, so there are limits to integration, the numbers a and b. So, our task is to find out where U is here, where dV is here. As U, we should take a function that has a good derivative. Obviously, we take U to be lnx. As a dV, what do we have? dV always has a multiplier of dx, so we have xdx left.
To apply the formula, we need to find dU. So, what is a differential? This is the derivative multiplied by dx, this is dx divided by x. Here (see video) to find V, V is restored by integration, this is one second x^2. Note that this is not an integral, in general, it is one of the primitive, not a family, but one of the primitive. So, let’s see what we have. So, we continue calculating the integral. We must multiply the functions U and V, lnx, and one second x^2. Let’s write down these multipliers so that it is easier to read (see the video). We have a definite integral, so we set the integration limits from 1 to 2.
Next, we write minus, the sign of the integral and VdU, one second x^2, dU is dx divided by x, from 1 to 2.Here, it is logical to perform the calculation immediately. So, one second of what should we multiply? The calculation at point 2 is 4ln2-0. Here, we will write: we take one second out of the sign of the integral, reduce it by x, leaving xdx from 1 to 2. We continue, perform the operation, and get 2ln2. Here is a table integral equal to x^2 divided by 2, so it will be minus one-fourth of x^2 from 1 to 2.Well, then we calculate x^2 from 1 to 2. Here, we rewrite 2ln2 minus one fourth, let’s calculate what this expression is. This is at the top point of 4, subtract 1, equal to 3. You will have to multiply by three, which means minus three fourths. That’s it, the calculation is complete. So, we applied the method of integration in parts.
So, again we see a definite integral. What is under the integral? What is the function? What is the type of integrals? Let’s see if the denominator is a square trinomial with a negative discriminant, and the numerator is a linear multiplier. Generally speaking, this is the simplest fraction of the third type, so we will use the same methods for calculating it as we did. So, we write out the denominator, select the full square, it turns out (x+1)^2+4. Make sure if everything is correct. So as a new variable, we take x+1. This means that the expression (x+1)^2+4 becomes t^2+4. We’ve already changed the denominator, haven’t we? What else do we need to find out?
That’s not all, we need another 2x-1, we need dx. What should we do with integration limits? Let’s take it one at a time. So, x we get is t-1. So dx and dt are equal, they differ by a constant. So, dx, it is already clear, we will replace with dt. What else will we do next? Let’s see, what will the numerator turn into? So, 2 multiplied by t-1, and -1 more. This is 2t-3 if we open the brackets. See, everything that relates to the integrand, we already know what to replace. Limits of integration. Here is an important point that you sometimes forget that in addition to changing the variable, the differential of the variable, it is necessary to replace the integration limits. These changes relate to the variable x. x was equal to 1, but we will have a new variable in this place. t is x+1, t is 2. At the top, x was equal to 2, t is one more, it will be equal to 3. That’s it, in my opinion, we can move on to the replacement.
So, the limits of integration have changed from 2 to 3, let’s see, in the numerator, this is 2t-3, in the denominator, t^2+4, dt. Further, we always do this when calculating integrals from the simplest fractions of this type, we divide the numerator by the denominator. So, the first term is 2tdt on t^2+4 from 2 to 3, and the second term is 3dt on t^2+4. Here again, we apply the method of adding a function under the sign of the differential. We know that there is a function and its derivative. Look, t^2+4, and 2t is the derivative. Look carefully. We find the derivative of the denominator, this will be 2t. So the numerator contains the differential of the denominator. This is the differential d(t^2+4), in the denominator we repeat the entry t^2+4 from 2 to 3. Here, it is logical to take 3 out, the integral dt divided by t^2+4. So, t varies from 2 to 3. Both of the resulting integrals are tabulated. So, the first integral is ln(t^2+4), the module can be omitted because the expression is always > 0, from 2 to 3, minus three second arctangents t divided by 2, where t changes from 2 to 3.
We have come to another difference from the indefinite integral. First, I remind you, we must not forget to change the limits of integration. This is the first difference from the indefinite integral. Second, now in an indefinite integral, we would return to the variable x, but in a definite integral, this should not be done. And in this case, the solution is even reduced. So, we calculate and immediately come to the answer. We calculate at point 3: 9+4, ln13-ln8-3/2(arctg3/2-arctg1). We can perform the last calculations. The difference of logarithms, we slightly reduce the entry, will be ln13/8-3/2arctg3/2+3П/8. The problem has been solved.
So, let’s turn to the last of today’s tasks. Again, there is no function here, as long as its differential. But we see the integrand contains e^x. What will we do? We multiply the numerator and denominator by e^x (see the video). x varies from 0 to 3. And now we introduce the new variable t=e^x. Look, in this case, dt is just e^xdx. This is what we have in the numerator. In the denominator, everything is also replaced perfectly. Do not forget the important point, replace the limits of integration. If x was 0, we assume that t is e^0, which is 1. At the top point, x was equal to 3, then t became equal to e^3. All the substitutions have been made, we write a new integral from 1 to e^3, in the numerator dt, in the denominator t(t+1). We have obtained the integral of a fractional-rational function.
If we know how to decompose the sum of the simplest fractions, we can do this. And I suggest performing identical transformations. So we had 1, we add t and subtract t, we write dt after the line, so t(t+1). We divide. What do we see? We divide 1+t by the denominator, 1/t is the first fraction, and we divide t by the denominator, it is 1/(t+1), dt, and e^3 here. We have already calculated quite a lot of integrals of this type, so I do not go into details, just write, here will be ln|t| - ln|t+1|. And don’t forget to write the integration limits, from 1 to e^3. We calculate. Again, a special feature – we do not return to the original variable. So, ln(e^3) is 3, we will immediately write ln1 is 0. That’s what we get. Here is a minus, let’s calculate, ln(e^3+1) - ln2. So, well, let's write down further, 3-ln(e^3+1)+ln2. In general, this is the answer we have come to.