Newton-Leibniz Rule

Newton-Leibniz's rule helps us to find the derivative of an integral, with respect to some variable, where the limits of integration are some functions of the same variable. While there are many variations of this rule (see note 1 below), we consider here one of the simplest forms.

The Theorem:

Let ` f(t) ` be a continuous integrable function and let `u(x)` and `v(x)` be some functions of x. Then the derivative of the integral of `f(t)` w.r.t. x is given by, 

` \frac{\text{d}}{\text{d}x}\int_{u(x)}^{v(x)} f(t) dt = f(v(x))\cdot v'(x) - f(u(x)) \cdot u'(x) `

Proof: The proof for this is very simple. Let g(t) be the anti-derivative of f(t). Then we can write, 

` \int_{u(x)}^{v(x)} f(t) dt = g(t) |_{u(x)}^{v(x)} = g(v(x)) - g(u(x)) `

Now taking the derivative using the chain rule, we get, 

` \frac{\text{d}}{\text{d}x} [g(v(x)) - g(u(x))] = g'(v(x)) \cdot v'(x) - g'(u(x)) \cdot u'(x)   `

However, by our assumption since` g ` is the anti-derivative of `f`, `f` is the derivative of `g`. This implies that ` g'(t) = f(t) ` for any `t`. Thus ` g'(v(x)) ` becomes ` f(v(x)) ` and ` g'(u(x)) ` becomes ` f(u(x)) `. The above then becomes, 

` f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x) `

The proof is thus complete. 

Applications to a Problem:

From a heuristic perspective, Newton-Leibniz helps us to eliminate the integral from a given expression. Now we shall take a very interesting problem requiring the use of the rule. The problem is as follows:

Prob. Find the curve that passes through the origin and is such that the area included between the curve, the ordinate, and the x-axis is twice the cube of that ordinate. 

Here we have to find the curve that satisfies the above conditions. To find any curve satisfying certain given equations we have to, in most cases, find the differential equation of the family of curves satisfying the conditions. 

Now we go into the problem. First of all, it is given that the area is twice the cube of the ordinate. i.e. at any point (x, y) under consideration, we must have, 

` Area = 2y^{3} `

Since area is always a positive quantity (see note 2 below), we must have ` 2y^{3} \geq 0 `, which implies ` y \geq 0 `. Therefore, whatever the curve, we only have to consider the part that has a positive ordinate.

For simplicity, we also take ` x \geq 0 `.

The Differential Equation:

Because we will be using the point (x, y) as the locus point, we use ` Y ` and ` X ` as the axis variables, i.e., now the curve is Y = F(X). This is to avoid confusion between the locus variables and the axis variables. From the diagram, the required equation is then,

` \int_{0}^{x} Y \text{d}X = 2y^{3} `

See that limits of integral are functions of ` x `, while the integrand is a function of `X`. Differentiating both side by `x`, we have, 

` \frac{\text{d}}{\text{d}x} \int_{0}^{x} Y \text{d}X = \frac{\text{d}}{\text{d}x} 2y^{3} `

Applying the Newton-Leibniz rule,

` Y(x) \cdot \frac{\text{d}}{\text{d}x}(x) - Y(0) \cdot frac{\text{d}}{\text{d}x}(0) = 6y^{2} \cdot frac{\text{d}y}{\text{d}x} `

 

` Y(x) frac{\text{d}}{\text{d}x}(x) = 6y^2 frac{\text{d}y}{\text{d}x} `

See that Y(x) = y, as the value of a curve at any point x is simply y. Thus, this further gets reduced to, 

` y = 6y^{2}frac{\text{d}y}{\text{d}x} `

This is the required differential equations of the family of curves satisfying the given conditions. We now have to solve this differential equation to find the curve. This is a simple variable separable form of differential equation. Solving this we have, 

` dx = 6y dy `

` \int \text{ }dx = \int \text{ }6y dy `

` x = 3y^{2} + c `

Since the curve passes through the origin (0, 0), c = 0. Thus, the equation of the curve is simply, 

$$ \boxed {y^2 = \frac{x}{3}} $$

The Second Case:

When `x \leq 0`, we have to change the integration limits to achieve a positive area.

As is seen from the diagram, the required equation in this case is, 

` \int_{x}^{0} Y \text{d}X = 2y^{3} `

Applying the same processes as above we shall arrive at, 

`0 - Y(x) frac{\text{d}}{\text{d}x}(x) = 6y^2 frac{\text{d}y}{\text{d}x} `

` y = - 6y^{2}frac{\text{d}y}{\text{d}x `

` dx = - 6y dy `

Integrating this we get,

` \int \text{ }dx = \int \text{ } -6y dy `

` x = - 3y^{2} + c `

Again c = 0, since curve passes through the origin (0, 0). Therefore, in this case, the curve is, 

$ \boxed{ y^2 = - \frac{x}{3} } $

The Final answer:

Combining the found curves for both the cases, ` x \leq 0` and `x \geq 0`, we have the following curve, 

$ \boxed{ f(x) = \begin{cases} y^2 = - \frac{x}{3} & \text{for } x \leq 0 \\ y^2 = \frac{x}{3} & \text {for } 0 \leq x \\  \end{cases} } $

This is the complete set of curves which satisfies the given conditions. Our solution is thus complete.

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Notes:

1. A more descriptive form is ` \frac{\text{d}}{\text{d}x}\int_{u(x)}^{v(x)} f(t,x) dt = \int_{u(x)}^{v(x)}\partial_{x} f(t,x) dt `, where f is a function of both x and t.

2. In Physics, area is often referred to as a vector quantity, where its direction of orientation is normal to the surface whose area is being considered. For a closed loop, the direction of the area vector can be found according to the right-hand thumb rule. Thus, when we transverse the loop in an anti-clockwise manner, the area vector is oriented perpendicularly upwards to the plane of the loop. In fact, even if the loop is not planar, it can be partitioned into small planar closed loops, where the addition of the area vectors of all such loops, will give us the net area vector of the loop. But mark that, in such a partition all the partitioned loops must add to give the net loop. 

Thus, when we say that the area is -5, what we mean is that the magnitude of the area of the surface is 5 and the direction of the normal of the surface is opposite to what we have considered a positive direction. In mathematics too such a direction of area is considered when dealing with closed domains, but in this question, since a particular orientation is not explicitly mentioned, it is to be assumed that it refers to the magnitude of area.