# Properties of Integrals and differentials

• andyrk
In summary: That is, h = h(x). So, y + h = y(x + h(x)) and y1 + h1 = y1(x + h1(x)). Therefore, ((y + h) - y) - ((y1 + h1) - y1) = y(x + h(x)) - y(x) - [y1(x + h1(x)) - y1(x)]. Now, if y and y1 are two different functions, then this expression is not equal to (y - y1) + h - h1. So, I should not have omitted the functions. Sorry. In summary, the integral is with respect to t so ex is a constant, and differentials are linear maps that

#### andyrk

I had a couple of questions.

1. Why does the integral ∫exf(t) dt transform to exf(t) dt? Shouldn't ex be a part of the integrand too?

2. Why is the difference dy - dy1 = d(y - y1)?

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andyrk said:
I had a couple of questions.

1. Why does the integral ∫exf(t) dt transform to exf(t) dt? Shouldn't ex be a part of the integrand too?

2. Why is the difference dy - dy1 = d(y - y1)?

1. The integral is with respect to t so ex is a constant.

2. Differentials are linear maps.

lavinia said:
2. Differentials are linear maps.
What does that mean?

A linear mapping is one such that f(x*y) = f(x)*f(y) and f(ax) = af(x), where * is some operation and a is a constant.

andyrk said:
What does that mean?

Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1

andyrk said:
What does that mean?

A function (also called a "map" or "mapping") has a domain and codomain that are sets. These sets need not be sets of real numbers. They need not be sets of any kind of numbers. They can be sets whose elements are very complicated things.

You can consider "taking the derivative" as function who domain and codomain are each sets of real valued functions of one real variable.

If we write "take the derivative with respect to $t$" in function notation we could write the result of doing this as $D_t(f(t))$ or just $D()$ when we are talking about real valued functions of one real variable.

The function $D()$ satisfies:
1) $D(f + g) = D(f) + D(g)$
2) $D(kf) = k D(f)$ when $k$ is a constant real number.

Those are the properties that define a "linear mapping" or "linear operator".

As an example of a function from a set of functions to a set of functions that is not linear, consider the function $H(f) = f^2$.

It's easy to get the impression that the word "linear" is always denotes something that has to do with a line (such as a "linear equation"). Often the word "linear" is used as an adjective without any implication that a straight line is involved. For example there are "linear operators", "linear differential equations" and the subject of "linear algebra". Sometimes a "linear" thing has indirect associations with a line. For example, in the case of the derivative operator, it produces a function that is interpreted as slopes of tangent lines.

As another example of an operator that is not linear consider $H(f) = af + b$ where $a$ and $b$ are constants and $b\ne 0$. It reminds us of the equation of a line, but it does not satisfy property 1) above.

Svein said:
Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1

Yeah! That was a quick and simple explanation..Thanks! :)

Svein said:
Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1
But wait. On second thought, d(y - y1) ≠ ((y + h) - (y1 + h1)) - (y - y1) because d(y - y1) = ((y - y1) + h) - (y - y1). How would you rearrange that to make it equal to dy - dy1?

andyrk said:
But wait. On second thought, d(y - y1) ≠ ((y + h) - (y1 + h1)) - (y - y1) because d(y - y1) = ((y - y1) + h) - (y - y1). How would you rearrange that to make it equal to dy - dy1?
It depends on how you view y1. If y1 is a constant, then your reasoning is correct. But if y and y1 are two different functions of x, say y(x) and y1(x), you must treat them as I did.

Svein said:
It depends on how you view y1. If y1 is a constant, then your reasoning is correct. But if y and y1 are two different functions of x, say y(x) and y1(x), you must treat them as I did.
They are two different functions. But why should that mean that we do it your way and not mine?

Is it because we look at dy as corresponding change in y when we change x? That is, dy by itself makes no sense, it is always accompanied by a change in x which further creates a change in y?

andyrk said:
That is, dy by itself makes no sense, it is always accompanied by a change in x which further creates a change in y?
Yes. I apologize for being a trifle sloppy with my notation. The correct (and long-winded) way is: $\Delta y(x) = y(x+\Delta x) -y(x)$ and the same for y1(x). Then $lim_{\Delta x\rightarrow 0}\frac{\Delta y(x)}{\Delta x}=y'(x) = \frac{dy}{dx}$ (and the same for y1). So, when we say y + h, we are in reality saying y(x + Δx), so h is a function of x.

## 1. What is the difference between an integral and a differential?

Integrals and differentials are both mathematical concepts used in calculus. Integrals represent the area under a curve, while differentials represent the change in a function. In other words, integrals are used to find the total accumulation of a quantity, while differentials are used to find the rate of change of a quantity.

## 2. What are the properties of integrals?

The properties of integrals include linearity, the fundamental theorem of calculus, and the power rule. Linearity means that the integral of a sum is equal to the sum of the integrals. The fundamental theorem of calculus states that the integral of a function is equal to the difference between its antiderivative evaluated at the upper and lower limits. The power rule states that the integral of x^n is equal to (x^(n+1))/(n+1).

## 3. How do you evaluate an integral?

There are several methods for evaluating integrals, including the power rule, substitution, and integration by parts. The specific method used depends on the complexity of the integral and the tools available. In some cases, integrals can also be evaluated numerically using techniques such as the trapezoidal rule or Simpson's rule.

## 4. Can integrals be used to solve real-world problems?

Yes, integrals have many real-world applications, such as calculating areas, volumes, and work done. For example, integrals can be used to find the distance traveled by an object given its velocity over time, or the amount of paint needed to cover a curved surface. They are also commonly used in physics, engineering, and economics.

## 5. What is the relationship between integrals and derivatives?

The relationship between integrals and derivatives is described by the fundamental theorem of calculus. This theorem states that the derivative of the integral of a function is equal to the original function. In other words, integration and differentiation are inverse operations. This relationship is a fundamental concept in calculus and is used to solve many problems in mathematics and science.