Differentiation of function has a method to solve, by limits
$$ \frac{d(f(x))}{dx} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
Is there any method by which we can solve integral without using antiderivative, like differentiation does?
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1I think you are misinterpreting the Fundamental Theorem of Calculus as the definition of an integral. The actual definition of the (Riemann) integral is a limit of sums. – Alborz Jan 01 '23 at 05:16
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2Riemann integrals certainly has a definition using areas of partitions – David Raveh Jan 01 '23 at 05:16
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But we can't integrate directly, like integral of (1+sin²x)^1/2.dx, but we can differentiate using limit formula. – Jan 01 '23 at 05:19
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1You can add the sums directly to get an approximation, that by choosing the width of your rectangles and adding more and more terms, can get arbitrarily close to the true area under the curve. No antiderivatives involved in this procedure. In this sense it is analogous to approximating the derivative by taking the limit for a small h and getting your approximations better and better for smaller and smaller h. – Alborz Jan 01 '23 at 05:24
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This may be helpful: https://math.stackexchange.com/questions/262397/functions-with-no-closed-form-derivative – David Raveh Jan 01 '23 at 05:27
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@Alborz Can you give me a example like integral of sinx.dx by this method? Because I don't know how will I calculate $$\lim_{h\to 0} h(\sum sinx)$$ where upper and lower limit are also not known – Jan 01 '23 at 05:37
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@Siddharth A rigorous definition is given here: https://en.wikipedia.org/wiki/Riemann_integral. This is the definition that is taught in real analysis; in most introductory calculus courses, it is done with areas of trapezoids: https://en.wikipedia.org/wiki/Trapezoidal_rule – David Raveh Jan 01 '23 at 06:23
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1Can you give me a example like integral of sinx.dx by this method? --- See Computing $\int_0^\pi \sin(x) ; dx$ using the definition. – Dave L. Renfro Jan 01 '23 at 08:06
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Thanks to all, I got it! – Jan 03 '23 at 06:03
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Similar question: Is there a formal definition for antiderivatives? – user170231 Jan 04 '23 at 21:08
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The short answer is "no". There is no mechanical way to integrate generally. Additionally, you will find that there are a lot of "basic" integrals that have no symbolic solutions. (Note that, technically, anything has a symbolic solution if you simply define a new symbol. However, integration doesn't have any closed set like differentiation does.)
Example: $\int \cos(x^2)\,dx$ does not have a solution, except by introducing a new symbol to represent its results (and, in fact, the "fresnel integral" function is often used for that one).
In mathematics, not everything can be mechanically determined. This comes as a surprise to many, but it is a fact of mathematics.
johnnyb
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As you said "fresnel integral", this too have been solved or proved by basic defination or using other integral or either should have its own definition. Right? – Jan 04 '23 at 19:57
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@Siddharth No. We can evaluate this integral over, say, $(0,\infty)$, but not in general. There's more to integration than finding antiderivatives, but we know for sure there are plenty of functions for which no "elementary" (=nice) antiderivative can be found – FShrike Jan 04 '23 at 19:59
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1It has a power series expansion, but nothing "great" in my eyes. There's not always a clean closed form for expressions - a fact one should accept when studying mathematics. – Sean Roberson Jan 04 '23 at 20:00
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@Siddharth What I always tell my class is that, there is nothing nonsensical about finding the area under $\cos(x^2)$, but that doesn't mean there is a function for it. However, we can define whatever function we wish. So, if the area under $\cos(x^2)$ is a sensical idea, then I can simply define a function, $Q(x)$, where $Q(x) = \int \cos(x^2),dx$. A lot of math is done this way - simply identifying pieces of interest is a step forward. – johnnyb Jan 05 '23 at 02:13