Calculus - Find Area of A Board By Using A Definite Integral & When is an Area Positive or Negative скачать в хорошем качестве

Calculus - Find Area of A Board By Using A Definite Integral & When is an Area Positive or Negative

In this video we will be discussing integration with limits in laymans terms… So I will first lead with a question we have a board that is 4 inches by 6 inches and has an area of 24 inches squared We have another board that is 4 inches by 3 inches and has an area of 12 inches squared If we were to place the smaller board ontop of the other how would we possibly find the area that is not covered by the smaller board??? That’s right we can take the area of the larger board and subtract the area of the smaller board to get 12 inches squared. Now lets put these 2 boards on a graph. The shared height of 4 inches will be what we set y equal to. Now how would be go about using integration or finding the area between the equation y =4 and the x axis to solve this problem? Well we have a constant of 4 that needs to be integrated between length x =6 and length x = 3 Integrating this using the constant rule we are left with 4 times x (x being the desired lengths) this is all for the limit between 6 and 3. Which means the area between x =6 and x =3 on the graph y = 4. So this is the same as our board example. When taking a limit we take the top value plugged into the equation minus the bottom value plugged into the equation. Plugging in our numbers we get a area of 12 inches squared. Now lets do the same problem but this time we place the boards vertical. How would we graph this? We would graph it with y=6 for the larger board and y = 3 for the smaller board. What would the integral look like for this one? Well we would take the integral of 6 length from x = 4 to x = 0 for the first board then subtract the second boards integral of 3 length from x =4 to x = 0 Completing the constant rule we are left with 6x -3x and this is all from the limit of x=4 to x=0 So plugging in all of our numbers we get the displayed equation that once again equals 12 inches square To help with your intuition on how area work for integrals. I have a graph of y=x and y=-x on this slide. Integrating this we get ½ X squared positive and negative respectably which is graphed to the right. This shows if the area between the equation line and the x axis is above the x axis it is positive area when using limits. So if the equation is in quadrant 1 and quadrant 2 the area is positive. This means that on the integral of x and negative x graph the equation line will be increasing in value If the area between equation line and x axis is below the x axis then it is negative area when using limits. So if the equation line is in quadrant 3 or quadrant 4 the area is negative. This means that on the integral of x and negative x graph the equation line will be decreasing in value Disclaimer These videos are intended for educational purposes only (students trying to pass a class) If you design or build something based off of these videos you do so at your own risk. I am not a professional engineer and this should not be considered engineering advice. Consult an engineer if you feel you may put someone at risk.