so this video is for those of you who might be taking physics with calculus and there's two fundamental areas of calculus that you need to understand derivatives and integration now just to review let's say if you want to find the derivative of x to the n with respect to x using a power rule it's equal to n times x raised to the n minus 1. so for instance let's say if you wish to find the derivative of x to the fifth power it's going to be five x to the five minus one which is five x to the fourth now some of you might already know this but for those of you who don't this is just a quick refresher so let's say if we want to find the derivative of x to the seventh power you could try this if you want to but it's going to be 7 x to the seven minus one which is seven x to the sixth power now what about this one what is the derivative of five x to the eighth power and the five is just a constant so this is five times the derivative of x to the eighth power and the derivative of x to the eighth power is eight x to the seventh power five times eight is forty so the final answer is 40 x to the seventh power and so that's a quick and simple way to find the derivative of a function now integration is basically the reverse the indefinite integral of x to the n is going to be x to the n plus one divided by that result and then plus the constant of integration anytime you differentiate a constant you get zero so when you integrate you're going to introduce a constant so let's put this into practice what is the anti-derivative or the indefinite integral of x cubed dx so this is going to be x to the 3 plus 1 divided by 3 plus 1 plus c so this ends up being x to the fourth divided by 4 plus c and so you can write that as 1 4 x to the fourth plus c now let's try another example let's find the indefinite integral of 7 x to the fifth power so for this example you could take the constant and move to the front so this is the same as seven times the integral of x to the fifth power dx now using the power rule for integration the anti-derivative of x to the fifth is x to the sixth divided by six and then plus c so our final answer is seven you can say seven over six times x to the 6 plus c now i want to talk about the relationship between derivatives and division with that of integration and multiplication so in physics sometimes you may have a simple formula to deal with other times the formula might involve calculus for instance work is the product of force and displacement now displacement could be in the x direction or it could be in the y direction in the x direction displacement you could simply call it x in the y direction it's really just y well technically it's the change in x or the change in y because displacement is the final position minus the initial position it could be along the x-axis or along the y-axis so whenever you see d it represents displacement now this formula should be applied if the force acted on an object that is the net force is constant what if the net force is not constant what if it varies based on the displacement of the object what if it changes in this case you need to use calculus whenever something is not constant you need to use calculus to get the answer notice that the force is being multiplied by the displacement it turns out that the work done by a variable force is the integral of that force where let's say that force is a function of the displacement along the x direction times dx now you're going to integrate it from a to b so a and b will be like an x value basically x1 will be the initial position x2 is your final position but what i want you to realize is that when you have a formula if you're multiplying two variables that formula is going to be associated with integration when you're dividing two variables that formula will be associated with differentiation or derivatives so let me give you another example average velocity which i'm going to represent like this is basically the change in position or the displacement over time so we're dividing displacement and time so this formula is going to be associated with derivatives it turns out that the instantaneous velocity as a function of time is the derivative of the position function with respect to time so as you can see you need to associate division with differentiation and associate multiplication with integration now let's see if you got the point that i'm trying to make let's start with a familiar formula in your physics textbook you'll see this formula final velocity is equal to the initial velocity plus acceleration now with this formula there's a lot of things we can do here's one thing i'm going to do i'm going to move the initial velocity to the other side so i have v final minus v initial is equal to 80. now this represents the change in velocity so i'm going to represent that as delta v where delta means change so i get this formula the change in velocity is equal to the acceleration multiplied by the time now let's save that formula we'll talk about it later now starting from the original formula that we have let's solve for a so i'm going to move this to this side so once again i have v final minus venus equals a t and then dividing by t i have that the acceleration is v final minus v initial over t and we know this is the change velocity so we could say that delta v over i guess you could say delta t this t is really delta t is equal to a so looking at these two formulas which one is associated with derivatives and which one is associated with integration so let's say if we're focused on this formula if the acceleration is constant you can just use this formula to calculate the change in velocity but let's say if the acceleration is not constant let's say it's a variable would you use integration or differentiation to calculate the change in velocity notice that we're multiplying a by t so that's going to be associated with integration thus the change in velocity will be the integral of the acceleration function where a depends on t from t1 to t2 from an initial time point to a final time point now let's think about what this means if we have a graph with the acceleration in the y direction and time in the x direction and let's say the acceleration is variable we can calculate the change in velocity of the object let's say from t1 to t2 or from a to b by finding the area under the curve now sometimes you could use geometry to find the area under the curve other times you have to use calculus for a shape like this but what you want to take from this is that the area under this curve highlighted in blue represents the change in velocity likewise if we go back to the other function where we said that work is the definite integral of the force when it's a function of x if we graph this where f is on the y axis the displacement is on the x-axis and let's say the force is a variable function if we integrate it from a let's say to b the area represented by the shaded region is equal to the work done by the net force so this is just not any force but it represents the sum of the forces acting on the object within that force so make sure you keep that in mind the area under the curve is associated with integration which is also associated with multiplication when you wish to find the area of an object let's say a rectangle it's a length times width you're multiplying the length and the width now let's talk about this other formula acceleration is the change in velocity divided by the change in time so here we're dealing with division so this tells us that acceleration is the derivative of velocity it's the instantaneous rate of change of the velocity function so if you wish to find the acceleration function all you need to do is differentiate the velocity function with respect to time now let's draw a graph now let's say this is the acceleration on the y-axis time on the x-axis actually let's use velocity so velocity on the y-axis time on the x-axis and let's say the velocity is just it varies it's not constant if you wish to find the acceleration from the velocity you need to find the slope of this tangent line so let's say we want to find the acceleration at let me use a different shape let's say at this point to find the acceleration at that point what you need to do is calculate the slope of the tangent line and you can find that using derivatives the tangent line is the line that touches the point that only it touches the curve at only one point the secant line touches the curve at two points so this is a secant line so keep that in mind a secant line touches the curve at two points a tangent line touches the curve at one point the secant line represents an average rate of change the tangent line represents an instantaneous rate of change if you wish to find a slope between two points you could use this formula the slope is y2 minus y1 over x2 minus x1 so this represents the average rate of change so if you wish to find the average acceleration you can use a similar formula but instead of y you're going to be using velocity v final minus v initial and instead of x you're gonna be using time t2 minus t1 so to calculate the average acceleration from velocity all you need is two points the initial velocity at the initial time and the final velocity at that final time now to calculate the instantaneous acceleration you need to use derivatives so the instantaneous acceleration is going to be the derivative of the velocity function at some time t so we should really write a of t here so this right here represents the instantaneous acceleration and this here represents the average acceleration