hey all welcome to homeschool hope everybody are doing good and welcome back to class 11 physics series today i am going to teach you the basic mathematic tools that is helpful to understand class 11 interval physics so usually these mathematical tools will help us in solving physics problems easily few concepts involve these mathematical tools so without having an idea about these tools if you study them you don't understand anything for example instantaneous velocity and instantaneous acceleration that we studied in motion in straight line chapter you know to solve the problems of those concepts this integration and differentiation is must so these mathematical tools makes our life more easier especially in case of physics and we do use them in chemistry also but its use is very much limited in chemistry but everywhere we use this integration and differentiation in physics there are so many derivations that involve these mathematical tools so let us understand few basic aspects of this integration and differentiation and also trigonometric functions you know trigonometry is the one which you have already studied in your basic classes so i am not going to trigonometry but trigonometry is also very important to understand certain concepts of physics right and i have already covered three chapters in class 11 physics video links are provided in the description you can go and check it out if you have not yet watched and coming back to the topic first let us understand why we use differentiation and integration in what situations we can use differentiation and integration what actually we mean by those tools so coming to differentiation so the word itself says differentiate so differentiation okay differentiate in the sense what to separate right so differentiate these things differentiate those things what do you mean by this it means that we have to separate okay we have to break into smaller and smaller units that's what is the another meaning of differentiation for example in your early ages you might have played this puzzles right so you will have this kind of stuffs where you need to arrange and give a particular shape or to give a particular picture right say imagine there are so many puzzles arranged and and it's a picture of something and if i break this into smaller and smaller pieces you know this is what we can call it as differentiation normally it's it's like a bigger puzzle uh you know the blocks were neatly arranged and that when i am breaking into smaller and smaller pieces right so this is what we do in differentiation right so here a big function a function which is big a big mathematical function okay this function we are breaking into small functions right it can be any function it can be polynomial function it can be algebraic function or it can be trigonometric function logarithmic function any function a bigger function normally we are breaking into smaller function okay so to do this we use a tool called differentiation right so using differentiation what we are doing we are doing a bigger function i mean we are actually breaking a bigger function into smaller function just like arranged puzzle how how you break into smaller and smaller pieces the same way a bigger mathematical function can be broken into smaller and smaller pieces okay so for that to do this we use a tool differentiation so this is a general basic idea and what is integration so integration is just opposite to this integration is a opposite tool for differentiation in integration what we do is too many small functions too many small functions we add to make a big function to make a big function okay so many small small functions are joined together uh to give a big function right so integration and differentiation are two opposite tools actually right so even the formulas that you learn is quite opposite so in differentiation big function uh you know broken down into smaller functions and in integration small functions you can join and make it a big function so this is this is what the meaning of differentiation and integration in a normal language okay but in mathematical language it is very much different the way you express them and what are the formulas what are the rules everything we have to study so now i gave you a basic idea now let us go to their mathematical meaning how we can express them how we can write them right so there are certain rules all of them we will learn first let us talk about differentiation okay so first we will discuss all the formulas of differentiation later we will go for integration usually differentiation is expressed something like this okay so this is the way of representing differentiation see we write it as d y by d x okay so here y and x y and x are variables so they are variables and they are related they are actually related with each other with each other through a function through a function okay and here one thing you must know x whatever you have written down in the denominator that is x is independent variable it is independent variable always okay and whereas y is dependent variable okay so remember x is always some independent variable both are variable but both are related to one another uh through a function okay so we normally write like y is equal to f of x so this is what the function a small function right so uh you know with differentiation in terms of differentiation we can write this as d y by d x okay so what do you mean by y is equal to f of x you you can represent this as d y by dx okay so where y and x both are variables both are related with each other through some function okay so this is the way we represent differentiation right and there are certain formulas that you have to understand in differentiation let us go to the formulas see guys first and foremost the first rule or the first basic formula is power rule power rule so when you have some polynomial function usually for example y is equal to u to the power n right you want to differentiate d y by d x that means you are different what is your y u to the power n that means you are differentiating u to the power n with respect to x so what do you mean by d y by d x you are actually differentiating y with respect to x okay it is just like rate of change of something okay so you are you are differentiating y with respect to x okay so what is y here something to the power n so when you have this the formula here is n comes outside it becomes u to the power n minus 1 you have to do into d u by d x okay so this is what the formula we will use whenever you have to differentiate something to the power n then n write it like this u n minus 1 you have to do into du by dx okay fine say uh one example here imagine if y is equal to you have x to the power n okay so then how do you differentiate d y by d x is equal to so what is y x to the power 1 you have to differentiate x to the power n with respect to x only okay so how do you write anything to the power n when you have n comes outside x n minus 1 into d x by d x right so usually d x by d x value is 1 only right so here your formula is n x to the power n minus 1 only because d x by d x is always 1 okay fine and ah let me give one example here so how do we use this particular formulas i mean this particular rule uh you have y is equal to x cube you have to differentiate so how do you differentiate the this d x cube by d x is equal to what is your n n is 3 right so n you will write what what is the format n into x to the power n minus 1 so what is n 3 into x to the power 3 minus 1 into d x by d x x only dx by dx okay so what is the answer here 3 x square is the answer dx by dx is 1 okay so 3 x square is the answer so this is how we use this power rule and another example example number 2 you have y is equal to 1 by x so how do you differentiate so this 1 by x you can write it as x to the power -1 so differentiation of x to the power minus 1 by d x is equal to what is the power minus 1 outside you write minus 1 into x to the power minus 1 minus 1 n minus 1 so n is what minus 1 so minus 1 minus 1 into dx by dx so you will get minus 1 x to the power minus 2 right so this is your answer so this one i can also write it as minus 1 by x square right so x to the power -2 it means it's a inverse inverse square right so i can write this as minus 1 by x square so differentiation of 1 by x is always minus 1 by x square right so this is where in different situations we can use this power rule and coming to next particular formula differentiation of constant is always equal to the zero okay imagine if y is equal to a a is some constant right so here d y by d x is equal to what is y d a that is some constant right so its value is always equal to zero for example example i'll show imagine y value is equal to 5 you have to differentiate 5 so differentiation of 5 with respect to x is always equal to 0 okay so this is how always keep in mind that differentiation of any constant is always equal to zero okay so and remember this is a very basic thing one must understand that is the power rule and now let us go for exponential rule and logarithmic rule so now exponential rule that means if you have a exponential function then what are the formulas we have to use for example okay so formula number one if you have y is equal to e power x then d y by d x that is that is d of e to the power x by dx is equal to e to the power x only okay so differentiation of e to the power x is always e to the power x itself so when you differentiate this with respect to 2 x then it is equal to e to the power x okay and now the second one if you have y is equal to e to the power y then differentiation of e to the power y with respect to x is equal to e to the power y into d y by d x okay so when you have uh when you have to differentiate e to the power y with respect to x then e to the power y into d y by d x okay so these are the two important uh formulas that you need to remember when you are uh differentiating exponents okay fine and coming though log rule log rule so how to differentiate logarithmic functions okay say uh if y is equal to lan x so lan is nothing but natural logarithm which is represented as ln okay so if y is equal to lan x then differentiation of lan x with respect to x is always 1 by x okay so differentiation of lan x is equal to 1 by x and now if you have to differentiate lan y okay with respect to x that is dx then the formula is 1 by y into d y by d x okay so these are few more important formulas that you need to remember so you will have their use in the later chapters okay fine and now let us understand the differentiation of trigonometric functions so coming to derivatives of trigonometric function derivatives or differentiation is one and the same okay so if you have to differentiate sine x with respect to x okay so d of sine x by dx is always equal to cos x okay so cos x is your answer differentiation of sine x is cos x okay and similarly differentiation of cos x by d x is equal to minus sine x okay so in this case it is minus sine x this is a formula differentiation of cos x is always equal to minus sine x and coming to differentiation of tan x with respect to x is sec square x okay and differentiation of differentiation of sec x with respect to x is equal to sec x into tan x sec x into tan x okay and coming to differentiation of cot x with respect to x is equal to minus cosec square x and differentiation of cosec x by dx is equal to minus cosi cosec x into cortex okay so these are some derivatives of trigonometric functions you have to remember along with that the basic idea on trigonometry like standard values of trigonometric function sine 0 is how much sine 90 is how much sine 45 sine 60 like that cos 0 cos 90 cos 45 you know all these values i think you must have studied in your basic classes right so now i am talking about the derivatives of trigonometric function so uh if you differentiate sine x the value is cos x and differentiation of cos x is minus sine x differentiation of tan x is x square x differentiation of sec x is uh sec x into tan x differentiation of cortex is minus cosec square x differentiation of cosec x is minus cosec x into cortex okay so these are some formulas that you need to remember fine and coming to uh addition rule say you have a particular equation y is equal to this much and you have to differentiate so what is the value of d y by d x okay so d y by d x you know you have to differentiate every single term okay so here differentiation of x by d x plus x to the power 4 differentiation of x to the power 4 with respect to dx plus differentiation of 1 by x by dx plus differentiation of 2 root x by dx plus differentiation of cos x by dx minus differentiation of 3 e to the power x with respect to dx you know like this when you have this particular equation where you have addition and subtraction terms so you have to apply differentiation for each and every term so we have learnt the formulas you just have to apply the formulas and solve this equation right so how much is d y by dx so differentiation of dx by dx is 1 1 plus and it is a power rule you have to apply right so here what is the value 4 into x to the power n minus 1 that is 4 minus 1 that means 4 x cube right so differentiation of this term is 4 x cube plus differentiation of 1 by x 1 by x means what guys you can write it as x minus 1 so that is minus 1 into minus 1 minus 1 that is minus 2 so how do we write this we can write it as minus 1 by x square right and then plus differentiation of 2 into root x 2 you bring it outside and differentiate root x okay so root x how do you write root x root x you can write it as x to the power 1 by 2 with respect to x you have to differentiate this okay so how do you differentiate this you will write it as 2 into 1 by 2 right into x to the power n is 1 by 2 minus 1 right so you will have minus 1 by 2 so how do you write this minus 1 by 2 x to the power minus 1 by 2 you can actually write it as 1 by root x right so the differentiation of this value is 2 into 1 by 2 into 1 by root x plus differentiation of cos x formula we have seen differentiation of cos x is minus sine x so minus sine x right minus differentiation of 3 into e to the power x so 3 outside differentiation of e to the power x is e to the power x only right so this is what the answer you will get here 2 2 gets cancelled so ultimately what is your equation turning to be 1 plus 4 x cube minus 1 by x square right plus 1 by root x minus sine x minus 3 e to the power x so dy by dx answer is this much right so this is how you have to use differentiation for such equation this this this is what is helpful for our physics this is how we apply differentiation in case of physics okay so we have seen various formulas of differentiation and now let us understand where exactly we can use this differentiation see guys this differentiation is used in many concepts of physics see one such example is understanding instantaneous velocity and instantaneous acceleration that we studied in motion in straight line see we can easily uh define instantaneous velocity and acceleration with the help of differentiation and we can easily solve the problems of those concepts with the help of differentiation and let's see how it can be done and let's let's understand instantaneous velocity as well as acceleration in terms of differentiation okay so for that just observe this particular plot where i have a graph of y versus the x okay so imagine i i got a curve something like this right so i didn't get a straight line i got a curve like this when we have to find a slope uh at a particular point you know at a particular point we draw a tangent isn't it so let me consider these are the two points of a curve right and uh you know what with respect to these points you know this is what uh the difference in x let me call this as delta x you know uh i'll draw a small triangle where slope of this triangle so how do you find the slope of this triangle this difference this is a small difference delta y and this is a small difference delta x okay so slope is equal to delta y by delta x okay so this is really a very very small difference right so delta y and a small difference delta x so slope is always equal to delta y divided by delta x which is nothing but tan theta right you see this and this particular thing can also be represented as d y by d x this small difference in y and small difference in x okay so delta means large difference whereas the moment you write small d it means that small difference okay so uh the this particular concept we can express in the form of differentiation right so we can we can uh write the slope as d y by d x okay so like this let me connect this to instantaneous velocity right so what is instantaneous velocity it is nothing but velocity at an instant of time velocity at an instant of time is called instantaneous velocity right say for example when you have a xt graph let me consider x t graph right so that is position time graph so this is a position and the horizontal axis is your time and when you have a straight line at an instant if you want to find the velocity you know slope of this line will give you value for velocity all that we have discussed in the video on graphs right okay so how do you find the slope of this uh point you know at this point if you want to find out the slope you know it is equal to the average velocity only right so when you have a straight line instantaneous velocity is equal to average velocity okay so the slope is same at any point and slope is common between any two points right so whereas if you get a curve something like this okay when your velocity is constantly changing when you have a changing velocity if at this particular time you know at this time i need to find the velocity then what do you do at this point you will draw a tangent right and the slope of the tangent slope of this tangent will give you instantaneous velocity and you see here it is a very very small change in distance this is a very small change in distance and this is a very small change in time right so here slope slope is nothing but your velocity instantaneous velocity okay instantaneous velocity is equal to what we can write delta x by delta t we will write whereas here uh we have to apply the limits right so providing limits delta t tends to zero okay so when delta t almost this this difference almost when it approaches to zero then delta x by delta t will be the value of slope and the value of slope is nothing but the instantaneous velocity right say this one we can also write it as dx by dt right see we can also define our instantaneous velocity okay so instantaneous velocity can be defined as dx by dt okay so small change in x that is in distance by small change in time okay so instantaneous velocity can be defined in terms of differentiation right so instantaneous velocity is nothing but dx by dt so if you have a particular position given in a question you can differentiate that you will get the value of instantaneous velocity okay so when we solve a problem you would understand better actually right and how do you find the instantaneous acceleration similarly you can also find uh instantaneous acceleration it can also be defined in terms of differentiation right so how do we define instantaneous acceleration it is dv by dt okay so r or d of what is v v is nothing but dx by dt right dx by dt see this is first derivative and this thing once again i am differentiating so second derivative i am doing okay so this will be d square x by dt square okay so acceleration can also be defined this way okay in terms of velocity if you want to define it is dv by dt right and in terms of position if you want to define acceleration it is d square x by dt square and this is called as second derivative okay so this was the first derivative right and and this factor two times we have differentiated so we call this as second derivative okay so acceleration in terms of differentiation you can define it this way or you can define it this way so these are the different formulas actually we will use it for solving problems okay so this is where the differentiation is helpful so whenever you want to find a very small difference a very small flow if you want to find a particular distance or velocity at the point at an instant of time then differentiation comes into picture okay so it makes our life more easier to solve the problems on instantaneous velocity as well as instantaneous acceleration okay so now let us look at one problem where with the help of differentiation we can solve it see guys look at this question the position of an object moving along x axis is given by x is equal to a plus bt square okay say the position is given by a particular equation that is x is equal to a plus bt square where a is equal to 8.5 meter b is equal to 2.5 meter per second square and t is measured in seconds what is its velocity at the t is equal to 0 seconds and t is equal to 2 seconds so they are asking velocity at this time instant and also velocity at this time instant so here only we can understand that they are asking instantaneous velocity at t is equal to 0 seconds and instantaneous velocity at t is equal to two seconds okay so at this time what is the velocity at this time what is the velocity they are asking okay so instantaneous velocity they are asking and the position of an object is given by this equation x means what position okay x is equal to this one so how do you do so we know instantaneous velocity is equal to dx by dt it can be represented as dx by dt with the help of differentiation okay so what is x here you have to differentiate what is x x is this equation differentiation of a plus b t square you have to do with respect to time so differentiate differentiation of a by dt plus differentiation of bt square by dt addition right differentiation of this term and differentiation of this term so differentiation of a a is a constant you see it's a constant so differentiation of constant is always 0 right and differentiation of bt square and and this b comes outside so differentiation of t square by dt okay so what is differentiation of t square by dt we have learnt a formula right differentiation of x square by dx is what 2 into this comes outside to uh or x to the power n how do you write n x to the power n minus 1 right so this was the formula same way here apply that 2 into t to the power 2 minus 1 so what is its value 2 t right so uh here what is the value let me write it here so your instantaneous velocity is equal to 0 plus b into 2t differentiation of t square is what 2t right so b into 2 t right meter per second so this much is your instantaneous velocity whereas here t is equal to time and now let us find that find out instantaneous velocity at t is equal to 0 so substitute t value here b what is b value b value is given that is 2.5 right into 2 into time is equal to 0 so your instantaneous velocity at t is equal to 0 is 0 only right so these many meter per second 2.5 into what is 2.5 b value into 2 into time time is 0 substitute t is equal to 0 in this form right so when t is equal to 0 your instantaneous velocity is also 0 like that your instantaneous velocity at t is equal to 2 seconds how much that is instantaneous velocity at 2 seconds is equal to b b value is how much 2.5 into 2 into time is how much 2 seconds right 2 seconds so how much do you get it is 2.5 into 4 right so 2.5 into 4 is how much it is 10 meter per second so 10 meter per second is your answer right so this is how you have to do a problem guys say x value they give what we have done by definition our instantaneous velocity is equal to dx by dt right so we have differentiated this equation okay so we got this much as instantaneous velocity and they asked you to calculate instantaneous velocity at time t is equal to zero so we have substituted value of time here we got instantaneous velocity at this time t is equal to zero similarly we substituted t is equal to two seconds so we got instantaneous velocity at a time t is equal to two when we substituted we got 10 meter per second as instantaneous velocity at a time t is equal to 2 seconds so this is how we can use differentiation to calculate uh instantaneous velocity as well as instantaneous acceleration also okay now let us see one more question that was asked in the previous year competitive examination see guys it's a very very simple question asked in one of the competitive examination okay so two cars p and q they are starting from same point okay in a same time and they are moving in a straight line and their positions are represented position of p is represented by this equation position of car q is represented by this equation at what time do the cars have same velocity so they are asking some instant of time at what instant of time actually the the cars have same velocity okay so they are telling at some time the cars have same velocity right so from this from this information i can say at some time at some time that time only we have to find out so at some time velocity of car p is equal to velocity of car q and let us try to find uh so these velocities are instantaneous velocities okay so let me find out so how do you find instantaneous velocity uh instantaneous velocity of car p is d x p by d t right is equal to d x cube right divided by d t okay xp is given by this equation xq is given by this equation you just have to differentiate these equations right so let me do it differentiation of what is xp eighty plus bt square okay divided by dt is equal to differentiation of x cube x cube is what differentiation of f t minus t square okay divided by d t okay and you see here uh how do you differentiate this say this is addition differentiation of this term plus differentiation of this term okay so uh here you know differentiation of dt sorry 80 by dt plus differentiation of bt square by dt you have to do okay so here differentiation of ft by dt minus differentiation of t square by dt i separated this is the equation addition right so differentiation of this term plus differentiation of this term now differentiation of 80 a is a constant guys okay so a is a constant so actually this is in the form of uv so here learn one one more rule for differentiation which i forgot to tell you initially initially i explained all the rules right include this rule also uv rule there is something called uv rule if y is equal to uv i mean when you have a multiplication of two variables right so how do you differentiate it is d y by d x is equal to d of u v by d x you have to do so differentiation of product is always done like this keep the first term constant u constant differentiate second term dv by dx u2 plus next v constant and differentiate the first term du by dx you have to do so this is the formula you have to apply in this case this is also in the form of uv a is u t is v okay so on applying this formula what do i do i'll keep a constant first and differentiation of t dt by dt by denominator also dt right so plus next i'll keep t constant so differentiation of d a by dt differentiation of a so a into a into dt by dt is 1 plus t into differentiation of any constant is 0 right so what is your value it is just a so you got a value ea right so differentiation of a t by d t is how much it is a right you got it here yay plus so similarly differentiation of bt square is how much b constant outside b 2 into t right so 2 bt so i'll write this way it is 2 bt okay so differentiation of a t plus b t square is how much a plus 2 b t is equal to differentiation of f t minus t square differentiation of f t is okay you can apply this formula uh uv uv rule you can apply here you will get f minus 2 t differentiation of t square is 2 t differentiation of x square formula do you remember so it is 2t okay so this is what you will get now adjust a matter of rearrangement i will just rearrange here so how i rearrange is f minus a is equal to this a i'm bringing here this minus 2 t i'm taking that side okay so a f minus f a i'm bringing here so f minus a is equal to this minus 2 t i am taking that side so you will have 2 t plus 2 bt right so f minus a is equal to here let me take 2t as common if i take 2t as common i get 1 plus b right so now f minus a by this one i'll shift here it comes in the denominator 1 plus b is equal to 2t so t is equal to how much so this 2 also if i shift this side it will be 1 by 2 f minus a 1 plus b is equal to t so what is your time time is this much okay see in which of the option it will match option a is this option b is this option c is this so definitely half into f minus a divided by uh 1 plus b or b plus 1 so option b is correct answer right see with the help of differentiation how we solve the problem right so this is how differentiation can be helpful in solving the problems relating to instantaneous velocity and instantaneous acceleration not only in these concepts it can be helpful in the further concepts which come in further chapters okay so this is all about differentiation now we will have some basic idea on integration i already told you integration is just an opposite tool to differentiation right so let's understand integration see guys as i already mentioned integration is a mathematical tool that helps to join small small pieces into a big piece we have small small pieces i want to join and make it make it to a bigger piece then i can always use this integration and to understand the concept of integration let me start with this graph imagine i have a graph of y versus x and suppose this is how i got a graph a straight line and if i ask of a question to find area covered under this graph what do you do you would just put one line perpendicular to the x-axis and you will try to find the area of this triangle now this is the shape of a triangle and we know a formula to find area of a triangle right half into base into height and using that we will find fine and now suppose if you have a graph something like this right i got a graph a line perpendicular to the x-axis and i asked you to find area covered under this graph then what we can do we just put a line perpendicular to the x axis and we will try to find this area so when you find the area of this shape so this is actually a rectangle right so we know to find the area of a rectangle that is length into breadth so we we measure the breadth this much so this divided by this difference right so this much we do it and we calculate the area right but suppose suppose if your graph is a curve so we have seen the curved graphs in straight line motion right so imagine if your graph is something like this and i asked you to find i asked you to find area under curve so this is a curve area under this curve how do you find say normally what did you do you just have drawn a line perpendicular to x axis you got some shape you got either a triangle or rectangle or some shape you had got previously using a formula you found the area but in this case how do you find the area of this uh you know this much so area under this curve to find its its not a shape right and its not a triangle it does not match with rectangle it is a curve but you need to find area under a curve then comes integration we can use integration in this case so normally suppose see i i divide this into this particular strip okay so i i'll divide entire uh you know graph into small small strips right so this is one strip this is another strip this is the third strip this is the fourth strip now it is a shape of a rectangle right say what is this this is y1 okay and this difference let it be delta x and this is y2 okay this difference is uh delta x okay so this is delta x1 this is delta x 2 this is delta x 3 this is delta x 4 right and this is my first rectangle and if i have to find its area how do i find i'll i'll find it as you know y 1 into delta x one so this is my second rectangle and i will find its area as y two so length into breadth right so this much is length that is y two is a length breadth is delta x right and this is my third rectangle and i will find the area as y 3 delta x 3 right fourth rectangle i will find its area as y 4 y 4 into delta x4 like this so the entire region i can divide into small small strips and individually i can find the area and later what i do if i want the total area if i want the total area okay so from uh a to b i want the total area from a to b what i'll do i can join these things y1 into x1 plus y2 into x2 plus y3 into x3 like that i can join the areas of small small rectangles and go on okay so sum of so many things how can we represent we can represent it as summation summation of y n n can be 1 2 3 4 and so on y n into delta x n i can write okay but we can't use the summation here when you when you have to add so many things we usually do summation this symbol we use right but we cannot use this symbol why because you see er is a curve see here this small areas uh this small regions area you are missing out it's a curve it's not a definite rectangle it's not a regular rectangle right somehow this reason this small regions this small regions area you're missing it out so you can't use a summation here okay so instead of that you know if you want to find area area from a to b this much area a to b if you want to find out instead of summation we can't use summation because small regions are missing out they are not counted right in our rectangle so to correct that we will use a factor called this this is the symbol for integration integration means what joining joining smaller and smaller pieces say say i i broke this entire region into small small rectangles each rectangle's area i can find and later i can join i can add all of them on doing that i get total area between a and b right so this joining joining small small rectangles joining off small small things we call it as integration so instead of summation we use integrate we integrate you know y y n into d x so why did i put dx because it's a small change in x right say i can't use delta also i have to use dx everywhere this is dx1 and this is dx2 and this is dx3 and this is dx4 right so delta means a big change but it's a very small rectangle it's a very small change right so for small change we use dx right so integration of yn into dx if you do when you integrate all of them you get you know a correct area under this graph so from where to where limits so integration from a to b okay so this is how you can show a limit right so if you do this if you do this you will get area under curve so usually whenever you want to find area covered under curve we always use the mathematical tool called integration okay so uh if you you are what's the meaning of this we are joining we are integrating all the areas from a to b okay from a to b we are joining all the areas of small small uh you know rectangles that is what the meaning of this right so this is where we actually use integration right say finding area under curve is very very important you have seen in motion in straight line right say how do you find the displacement from vt graph it is by finding area under curve right so in many of the concepts we keep using this integration so let us learn few formulas that comes under integration okay so with this formula you can easily solve many problems in physics so let's understand the basic formulas of integration so you guys the first basic formula so this is really very important most of the times we use this formula if y is equal to x to the power of n so integration of y into dx is equal to what is the value of what integration of x to the power n into dx integration of x to the power n into dx is equal to x to the power n plus 1 divided by n plus 1 plus c this c is called integration constant so why we have to introduce when we have to introduce this integration constant when you don't have any limits here when you no need to do from where to where you have to integrate then we use a constant called integration constant when you have a limits here from where to where exactly if you mention then we will not use this c remember okay so this is actually called as indefinite integration so you don't know from where to where you are integrating okay so indefinite integration when you do indefinite integration then integration constant you have to compulsorily mention okay fine say for example this formula i will use i i'll show you for example uh your y is equal to x square so how do you integrate integration of y into dx is equal to integration of x square into dx right so what is the formula integration of x to the power of n is equal to x to the power of n plus 1 divided by n plus 1 so x to the power 2 plus 1 divided by 2 plus 1 plus c that means x to the power 3 divided by 3 plus c so this is your answer okay so similarly example number 2 example number 2 okay so uh if you want to integrate integration of uh x cube into dx is equal to if you want to integrate x cube so how do you do it you will do x to the power 3 plus 1 divided by 3 plus 1 plus c don't forget to put c because it is a indefinite integration so your answer is x to the power 4 divided by 4 plus c okay so similarly example number 3 so integration of root x into dx okay so this root can be written as x to the power 1 by 2 right into dx right so how do you uh integrate here x to the power 1 by 2 plus 1 divided by 1 by 2 plus 1 plus c you have to do okay so this is how we can use this particular formula and remember this formula cannot be used when n is equal to minus 1 okay so this formula you cannot use the integration of x to the power minus 1 okay into dx when you have this usually if you apply this formula how do you calculate x to the power minus 1 plus 1 uh here minus 1 plus 1 that means 0 anything to the power 0 is undefined right so you cannot use this formula when you have n is equal to minus 1 for all other numbers you can use this formula that is integration of x to the power anything is x to the power n is x to the power n plus 1 divided by n plus 1 okay so you can observe these examples how we have used this formula right so now let's go for formula number 2 if y is equal to 1 by x okay so how do you integrate integration of y into dx is equal to integration of 1 by x into dx so integration of 1 by x into dx is always equal to log x okay log e to the power x okay so do you remember uh differentiation of log x was 1 by x so that's the reason we tell integration is the opposite of differentiation okay so differentiation of this term was equal to 1 by x whereas integration of 1 by x is log x okay so 1 by x is nothing but what integration of x to the power minus 1 into dx so that means when n value is minus 1 i told you in the previous case n value should not be minus 1 if n value is minus 1 it is nothing but 1 by x so integration of 1 by x value is log e to the power x okay so this is really very important now third one what is integration of just dx okay integration of just dx is x okay remember integration of just dx is x 5 and the fourth formula imagine if you have a limits now let us start with x plus c see this c you shouldn't forget because it is undefined integration you don't know from where to where you have to integrate then in that case don't forget to put the integration constant see now let us do a definite integration guys integration of x square dx okay from the limit 2 to 3 okay see now it is definite you just apply a normal formula what is integration of x square dot dx it is x to the power 2 plus 1 divided by 2 plus 1 limit 2 to 3 so you will get this x to the power 3 divided by 3 limits 2 to 3 so till here it is common now you have to apply this limits i mean you have to use these limits see this is a lower limit and this is a upper limit okay in the place of x substitute 3 okay so 3 cube divided by 3 minus in the place of x substitute lower limit that is 2 cube divided by 3 that's it so what we have done upper limit substitute in the place of x right minus in the place of x substitute lower limit so what is your answer it is a 3 cube 3 cube is how much 27 by 3 minus this is 8 by 3 if you do you will get the answer okay so whenever you have a limits this is how you have to do right so this is our fourth uh case that you need to remember we keep using in various problems and then coming to a fifth formula integration of e to the power x into dx answer is e to the power x only plus c okay when there is no limits integration constant e to the power x plus c okay and now you see a sixth case integration of 5 x square into d x see some constant is there so what you have to do take the constant outside and now you integrate you integrate integration of x square is what x cube 2 plus 1 is 3 divided by 3 okay so plus c right so whenever you have a constant uh when you have a product of constant and uh your variable then bring the constant outside then integrate this with the help of a formula plus c right so these are the six basic formulas or the six basic rules you must know as of now okay and there are so many formulas you have almost around two to three chapters in class 12 mathematics on integration you will come across many formulas you can use trigonometric functions here and everything it's it's a very very vast topic but this is very very basic thing that i told you and as you study the concepts various concepts in physics uh we we keep telling you uh the various formulas of integration okay so when situation demands we normally teach how to use various integration tools for solving a particular problem in physics right so i have just given you a general idea on differentiation and integration and a very basic formulas which we normally keep using in physics so that's all about today's topic hope you had got enough benefit out of this and in my next video i will start the chapter four that is motion in plain so to learn motion in plane learning vectors is very very important so in my next video i will start with vectors all about vectors right so till then keep learning and take care see you in the next video [Music]