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anantchowdhary
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Is there a mathematical proof that can prove that the integral is the antiderivative?
umm...i tried this out..i Followed pretty much of itHallsofIvy said:A little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely "below" the graph. Let An be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that [itex]A_n \le A[/itex] where A is the area of that region (the "area below the graph").
Now do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking An to be the total area of all those rectangles, we must have [itex]A \le A^n[/itex].
That is, for all n, [itex]A_n \le A \le A^n[/itex]
If f HAS an integral (if f is integrable) then, by definition, the limits of An and An must be the same- and equal to the integral of f from a to b. Since A is always "trapped" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.
That's pretty much the proof given in any Calculus book.
mathwonk said:suppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,
i.e.f(x)h < A(x+h)-A(x) < f(x+h)h.
satisfy yourself of this by drawing a picture.
now the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.
by the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).
now in creasing is not n eeded ubt makes it easier.
An integral as an antiderivative is a mathematical concept that describes the relationship between the derivative and the integral of a function. It states that the integral of a function is the antiderivative of its derivative.
The integral as an antiderivative is used in various mathematical applications, such as calculating areas under curves, finding volumes of solids, and solving differential equations. It is also an essential tool in calculus and helps in understanding the behavior of functions.
The proof of the integral as an antiderivative involves using the fundamental theorem of calculus, which states that the definite integral of a function is equal to the difference of its antiderivative evaluated at the upper and lower limits of integration. This theorem is then applied to show that the derivative of the integral of a function is equal to the original function.
There are some limitations to the integral as an antiderivative, such as the function must be continuous and defined on the given interval. Also, certain functions, such as those with vertical asymptotes, do not have antiderivatives.
Understanding the integral as an antiderivative is crucial in many fields of science and engineering, as it allows for the calculation of various quantities and helps in solving equations. It also provides a deeper understanding of the relationship between derivatives and integrals, which is fundamental in calculus.