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I took Calc I course and currently studying Calc II. I'm pretty good at integration and differentiation in terms of mathematically solving equations, but I don't truly understand the concepts.

When I apply integration rules to a solve a problem it just feels like magic to me, so I don't really get what's going on. For instance, if try to a solve an unfamiliar application problem I won't really be able to solve it. I feel like I'm just a useless calculator. At the beginning, When I started to learn about differentiation I had a pretty good understanding like finding the instantaneous rate of change, and understood the relation distance => velocity => acceleration (and why it works that way). Now I lost track.

I want to be able to use these concepts creatively in real-world applications. Can you suggest a book or a give me an advice on how to truly understand these concepts?

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    Were you given any _definition_ of definite integrals? ​ If yes, do you still remember it? ​ If yes to both, what is it? ​ (There are several possibilities.) ​ ​ ​ ​ – 2017-02-07
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    Maybe useful the releted chapters of : Morris Kline, [Mathematics and the Physical World](https://books.google.it/books?id=DeSzakR5LKoC&pg=PA363) (1959). – 2017-02-07
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    Maybe also : Richard Courant & Herbert Robbins & Ian Stewart, [What is mathematics ? An Elementary Approach to Ideas and Methods](https://books.google.it/books?id=_kYBqLc5QoQC&pg=PA400) (2nd ed., 1996). – 2017-02-07
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    @someone: $\int_0^{10}(5-x)\,dx=0$. What area is zero? – 2017-02-07
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    @MartinArgerami 5x-x^2/2 solving from 10 to 0 would be (0-0) = 0 I guess they would cancel each other over this interval. I tried to graph it there is a positive area from 0 to 5 and a negative area from 5 to 10 with the same size so the result is zero. – 2017-02-07
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    Was "area over an interval" actually how they "defined" it? ​ If yes, did they give a definition for area? ​ ​ ​ ​ – 2017-02-07
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    See [Fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning) : from the derivative function (plotted gives the rate of change of the "magnitude") you integrate and recover the "original" function. – 2017-02-07
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    @RickyDemer it is "the interval of integration" and the definition is $$ \int_{a}^{b} f(x) dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x$$ – 2017-02-07
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    @someone: and what would a "negative area" be? Can you show me an example of a square with an area of $-1$ m$^2$? – 2017-02-07
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    @someone : ​ That's the one which should make understanding the fundamental theorem of calculus easiest. ​ ​ ​ ​ – 2017-02-07
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    @MartinArgerami a negative area is an area below the x-axis. the better term would be is the area under the curve. If the curve is below the x-axis like from 0 to 5 it would be negative if above the x-axis like 5 to 10 it's positive in your example of integrating (5-x). So the $-1 m^2$ square would be the area under a curve below the x-axis. – 2017-02-07
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    @MartinArgerami we only take mod while doing them, so actually, there is modulus in x-5. so we have to split the integral. you know what I mean? – 2017-02-07
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    @someone: you can probably tell I'm trying to challenge your notions. The idea that the value of an area changes depending on position is ridiculous, and would make geometry useless. In particular, you seem to be claiming that the area of a square is not the square of the side: what would it be, then? – 2017-02-07
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    @PushkarSoni: I have no idea what you are talking about. Where is "mod" in this:$$\int_0^{10}(5-x)\,dx=0. $$ – 2017-02-07
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    $\int_0^{10}|(5-x)|\,dx=0$ – 2017-02-07
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    @MauroALLEGRANZA Thank you. I will check all of these. – 2017-02-07
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    Lol, I'm a Highschool student. though I have done lots of integration problems, even some advanced ones, as my board requires. so I have to do it anyway, at first it was boring, but now its one of the best things I know, it develops thinking out of the box to solve problems out of the box. It's really fun. – 2017-02-07
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    @PushkarSoni: that equality, with the absolute value, is not true. So I still cannot see your point. – 2017-02-07
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    @someone: I will, later today. – 2017-02-07
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    we don't have negative areas. – 2017-02-07
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    so, how'd you calculate area for a simple function like cos?given an interval, you've to break it. – 2017-02-07

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I really suggest PatrickJMT. He has a lot of categorized and very helpful videos and can help fill in gaps in knowledge so you have a good foundation. It certainly helped get through IB maths HL Calculus - options and (first year) University Calculus.

Applications, you can look at optimization, and a lot of physics and modeling aspects depending on your interests. Related rates may give you more insight and problems with volume, area etc.