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Ever since I was introduced to integrals, I've had trouble relating the idea of Riemann sums (or definite integrals) with derivatives. Now taking multivariable calculus, I'm trying to build a more intuitive understanding of the ideas behind this branch of math, and this is one of my biggest hurdles in the subject.

I know it has to do with the first Fundamental Theorem of Calculus, but so far nothing on the internet seems to explain it in simple terms. Could somebody help me out? Thank you all in advance.

(By the way, a few weeks ago I happened to ask for intuition behind differentiation rules, and I couldn't be more grateful for all the responses I received in this community. I hope to, eventually, give something back in return. But for that I need to keep studying math xD)

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  • $\begingroup$ I find velocity gives a reasonable intuition: the distance travelled in time $\delta t$ is roughly $v(t)\, \delta t$ so adding up a lot of these slices for the distance travelled from times $t_1$ to $t_2$ is reasonably given by $\displaystyle s(t_2)-s(t_1)= \int_{t_1}^{t_2} v(t) \, dt$. Meanwhile the velocity is the rate of change of distance, roughly $\dfrac{s(t+\delta t)-s(t)}{\delta t}$, so $v(t)=\dfrac{ds(t)}{dt}$ and it all fits. $\endgroup$ – Henry Nov 3 '15 at 9:04
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Intuitively, the fundamental theorem of calculus is saying that the total is the sum of its parts.

$$F(b)-F(a)=\int_a^bf(x)dx$$

Here $f$ is the derivative of $F$.

If I start at $x=a$ and move an infinitesimal amount, $dx$, then my change in $F$ will be $dy=F'(a)dx=f(a)dx$.

In general, if I am at $x\in[a,b]$ and move by $dx$ the change in $F$ will be $dy=F'(x)dx=f(x)dx$.

So the integral is saying that if you start at $F(a)$ and add up all the changes along the path from $a$ to $b$ you will end up at $$F(b)=F(a)+\int_a^bf(x)dx$$

Subtracting $F(a)$ from both sides yields the original formula.

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