I am a high school student who has recently started calculus (all my previous calculus experience is mostly in physics calculating moment of inertia and other such stuff). The way I used to view the integral was as a sum of the values the function which I am integrating takes up over the whole domain with an infinitely small increment in the function's input value. But the problem I am facing is this- if I integrate sin x from 0 to pi/2, at least at two points it takes a value greater than 0.5. But the integration (as done in all standard places) comes to 1. What am I understanding wrong about what integration means itself?
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$\begingroup$ In integration we are considering the area underneath the curve. So for each function value you should multiply by the width of the rectangle below the curve. $\endgroup$– EffMar 24, 2017 at 3:35
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$\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$– GNUSupporter 8964民主女神 地下教會Mar 24, 2017 at 8:31
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$\begingroup$ How the hell doesn't this have "context"? Why is it getting downvoted? $\endgroup$– MathematicsStudent1122Mar 24, 2017 at 8:36
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$\begingroup$ Each term in your sum should be weighted by the width of a tiny interval. $\endgroup$– littleOMar 24, 2017 at 8:41
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3 Answers
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I like to think of it like this: $\frac{1}{b-a} \int_a^b f(x)dx$ is the average value of $f$ on the interval $[a,b]$. I find it much easier and more intuitive to think about averages and the 'average value' of some function. Note this generalizes finite averages. So, $\int_a^b f(x)dx$ is the average value of $f$ times the length of the interval, which is the area under the curve of $f$.
So $\int_0^{\frac{\pi}{2}} \sin(x)dx = 1$ means that the average value of $\sin(x)$ on $[0,\frac{\pi}{2}]$ is $\frac{2}{\pi}$.
answered Mar 24, 2017 at 3:34
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$\begingroup$ But am i not finding the total sum? So even if i do it numerically lets say for each degree shouldn't I get approximately the same result? $\endgroup$– AryaMar 24, 2017 at 4:09
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1$\begingroup$ You are not finding the total sum. You are finding a weighted sum in some sense. You are taking the sum over small elements times the lengths of those segments. These are exactly what Riemann sums are, sums of the form $\sum f(y_i)(x_i-x_{i-1})$. We get the integral when the segments get smaller and smaller. $\endgroup$ Mar 24, 2017 at 4:25
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2$\begingroup$ Nah thank you. I have finally understood what you meant. sin x does take up the values greater than 0.5 but the input value,that infinitesmal increment value by which we multiply is so small that the total sum doesn't increase by much. Thanks a lot for your answer:) $\endgroup$– AryaMar 24, 2017 at 4:28
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Integration gets the area underneath a curve.
You can have an very long line with an area of 0, right?
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$\begingroup$ Yes you can. But sin x only takes up the value 0 from 0 to pi/2 once. $\endgroup$– AryaMar 24, 2017 at 4:07
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$\begingroup$ @Arya What's the problem? 1+1 = 2, yet there was never anything above a 1 in the sum. $\endgroup$ Mar 24, 2017 at 4:11
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"What does integration mean?" is a really broad question. I'll give you an answer that might help give you some insight into the physics that you've learned.
"What does integration mean?". Answer: adding up "small" pieces.
You've seen this with the limit definition of the (Riemann) integral, adding up the areas of thin rectangles. You'll add up small "pieces" to get integral formulas for the volume and surface area of a solid of revolution and the length of a curve.
In physics you've probably saw this already when you calculate the moment of intertia of an object by splitting it into small chunks of mass $dm$ and then adding up the moment of inertia for each piece: $I = \displaystyle \int r^2 dm$.
