Solving Indefinite Integration Equation

Solving Indefinite Integration Equation

A. We will find y, from the known statement of dy over dx equal four times x square. The first way to get y is that we should multiply each sides with dx. So that, we get dx times dy over dx equal four times x square times dx. Moreover, it is became dy equal four times x square times dx. Then, to get y, we should integrate each sides. So that, indefinite integral of dy equal indefinite integral of four times x square times dx. In addition, the result is y equal four third times x cube plus constant. Therefore, the solving problem is y equal four third times x cube plus constant.
B. The next videos explain about adding problems. Here are some of the examples.
1. We will find x where x minus five equals three. The first step is adding each sides with a certain number that can cause x equal some number. Here, we can add five to each side. Then, the operation will be x minus five plus five equal three plus five. So that, we get x equal eight. The solution is eight.
2. We will find a where seven equals four times “a” minus one. First, we should add a number (one) to each side. The operation will be seven plus one equal four times a minus one plus one. Then, we get eight equal four times “a”. The next step is multiply each side so that we can get some number equal “a”. Here, we can multiply each other with a number which is the opposite of coefficient of “a”. Then we get eight times a quarter equal a quarter times four times a. In addition, the result is two equal a. So, the solution is two.
3. We will find x where two third times x equal eight. We should multiply each other with the opposite of two third which is half of three. (Why? The reason can be read in problem number two). Then the operation will be half of three times two third times x equal eight times half of three. It became x equal twelve. Therefore, the solution of this problem is twelve
4. We will find x where five minus two times x equal three times x plus one. The first step is eliminating one of the variables in one side. For example we may eliminate x in the right side, so that we have to subtract three times x in each sides. We get five minus five times x equal one. The second step is subtracting them with five. So that we get five minus five minus five times x equal one minus five. It gives negative five times x equal negative four. The third step is multiply them with negative one fifth. It leads negative one fifth times negative five times x equal negative one fifth times negative four. And the result is x equal four fifth. So, the solution is four fifth.
5. We will find m where three minus five times open bracket two m minus five close brackets equal negative two. The first step is open the brackets by multiply five with the number in brackets. So we get three minus ten times m plus twenty five equal negative two. Then the operation becomes negative ten times m plus twenty-eight equal negative two. Then the next step is adding each side with twenty-eight. It becomes negative ten times m equal negative thirty. Then the last step is dividing them by negative ten. Therefore, we get m equal three. So, the solution is three.

C. The next interesting video is about logarithm. We will proof one of the character of logarithm.
Logarithm base x of A plus logarithm base x of B equal logarithm base x of A times B. Then, suppose that logarithm base x of A is l and it is equivalent with x raise to the l equal A. Next, suppose that logarithm base x of B as m and we get x raise to the m equal B. Then we get logarithm base x of A over B equal n. It becomes x raise to n equal A over B which is equal x raise to l over x raise to m equal x raise to open bracket l minus m close bracket. While n is l minus m. So that is proved that logarithm base x of A over B is logarithm base x of A minus logarithm base x of B.