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Because we usually deal with continuous functions in physics. For instance, we know that internal energy of a body is continuous. I.e. if a body has internal energy of $U_1$ and $U_2$ in two different states and $U_2\gt U_1$ then certainly there are infinite states between 1 and 2 so that internal energy of the body is between $U_1$ and $U_2$. In other words,...

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We don't always need calculus to obtain a general form: static problems (a weight sitting stationary on a table, for instance) often don't require calculus. Calculus is therefore typically necessary where change occurs. Let's say a physical, observable quantity $u$ (distance, pressure or whatnot) changes in function of an observable parameter $t$ (time, ...

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We usually know the expression or "formula" for a small part, and this formula takes different values across parts, so one easy way to solve the problem, if this is the case, is by adding together all the mini formulas which is what an integral does. For instance you want to calculate the area inside a circle of radius R. How do you start? Well, if you ...

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The displacement is equal to the area under a velocity time graph so in your example the change in displacement $dx$ in a time $dt$ is equal to $v\; dt$ with $v=\frac 3 x$ So $dx = v\; dt = \frac 3 x \; dt \Rightarrow x\; dx= 3\; dt$ and then you do the integration.

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$\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ becomes $\dfrac{v-v_{0}}{2}$ Hence; $\Delta x =\dfrac{v-v_{0}}{2}t=\dfrac{1}{2}\dfrac{v-v_{0}}{t}t^{2}=\dfrac{1}{2}at^{2}$

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This is a trivial kinematic deduction. \begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align} Integrating this, we would get $$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6}... 3 Indeed your first suggestion is wrong : \Delta x = v_o t + gt  Instead it should be  \Delta x = v_o t + gt^{2} (You can recheck it) Where you are wrong is here: According to your question v is the final velocity since (v=v_{0}+gt) So \Delta x\neq vt but instead it should be \Delta x =v_{average}\times t In uniform acceleration v_{average} ... 1 Let's take the first equation of motion which is : $$v=u+at$$ Integrate this equation to get: $$\int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt$$ this gives: $$x=ut+\frac{1}{2}at^2+x_0$$ The integration constant can be done away by putting the proper limits on x.(Assuming the ... 0 I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong:  \Delta x = v_o t + gt . A unit analysis will show you why:$$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds meters \not= meters + \frac{meters}{seconds}$$And we can see that the assumption is simply not true ... 1 There is no relation between the area being two-dimensional in your graph and what it means. For example, consider you make the x-axis as an indicator of the temperature, so what is the difference between x= 5 K and x = 8 K? of course it's  \Delta x = 3K . Now, isn't x a one-dimensional quantity? So the graph gains its meaning from you not from ... 0 I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement , which are 1 dimensional quantities. You are right. Area is a 2 dimensional quantity but what you have missed out is that you didn't use dimensional analysis properly. The dimensions of velocity is [LT^{-1}]. So if you multiply velocity with time, you get the ... 0 If we consider the velocity-time graph area under the whole line is the distance. For example, Now look at the second part of the diagram (rectangular, dark blue)$$v=\frac{dx}{dt}\rightarrow\int{vdt}=\int{dx}\rightarrow x=vt$$But the first part (triangle), we should consider acceleration$$a=\frac{d^2x}{d^2t}=\frac{dv}{dt}\rightarrow\int{adt}=\int{dv}\...

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The interpretation of the area under a curve, depending upon the curve, will vary. If it is a Velocity v. Time Graph, the area from a given time to another time, will be the distance traveled between those times. If it is an Acceleration v. Time Graph, the area from one time to another, will be the change in velocity of the object between those two times. ...

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The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.

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It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$. Remember that the integral is the sum of all the products $f(x)\;\text{ times } \;dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.

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Whenever you see a function that looks like: $$y = \tfrac{1}{2}kx^2$$ there's a good chance it came from integrating the function: $$\frac{dy}{dx} = kx$$ For example your distance function comes from integrating the velocity $v = at$: $$y(t) = \int v\,dt = \int at\,dt = \tfrac{1}{2}at^2$$ The spring energy function comes from integrating the ...

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You won't be able to find the potential because of the presence of drag and the forcing term. In other words, this is a non-conservative field and therefore it is impossible to define a potential. Luckily, though, you don't need that because you have an explicit expression of the acceleration and that's all you need to use the Verlet or velocity Verlet ...

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