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So to my knowledge a definite integral's significance is how it shows the "intensity" or area under the curve for a function.

However, I am confused then why the definite integral for x from 0 to 1 = 1/2 is it representing the area under the curve with the implicitly assumed base/restriction being the x axis? because the amount of area under the line/function x extends quite far if you go below the x axis from 0 to 1.

However, I am probably just misunderstanding definite integrals and/or the function we integrate?

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Intuitive idea:

Split the area between the curve and the $x-axis$ into thin vertical rectangular strips of width $\Delta x$. For any $x$ in the domain, as you move from $x$ to $x+\Delta x$, the value of the function will change from $f(x)$ to $f(x + \Delta x)$.

Let the height of the strip covering the width $x$ to $x+ \Delta x$ be $f(x)$. Then, the area of each strip is given by $f(x)\Delta x$. It is known that taking the limit of the strip's width approaching zero over an interval gives the value of the area under the curve or the definite integral.

Notice that the height of the strips are $f(x)$, which is the distance between $(x,f(x))$ and $(x,0)$.

This is where the base has been assumed to be the $x-axis$. Suppose you wish the base to be $y=k$, then, the height of each strip would become $f(x)-k$, and the corresponding integral will give the area between the curve and the chosen base.

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Shade the area between the function and the $x$ axis. The area that is above the x axis counts as positive and the area below the x axis counts as negative. The sum of all the area in these regions on the interval of integration is the integral.

This is why $\int_{-1}^1 x^3dx=0$, the LHS is negative and the RHS is positive and they cancel out.

You can take $\int |f(x)|dx$ to get the area between the x axis and the function with all of it counted as positive.

Algebraically, an integral is a "more than infinite" sum. Technically, it's a sum of uncountably many things, but each of them are infinitely small. This balances out to get a finite number (often times). You can think of an integral as adding up an infinitely thin slice going from the x axis to the graph of the function. This picture explains why some area is positive and some area is negative.

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  • $\begingroup$ It is never the sum of uncountably many things, whether you use the Riemann integral or the Lebesgue integral or some other integral. $\endgroup$ – user21820 Feb 5 '16 at 8:02

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