I know that if we start with an original function and take one derivative, we get another function. If we take the integral of that new function we get the original function back. So I see how they are opposite operations but I don't see how they are opposite in the sense that the integral is supposed to be the area under the curve and the derivative is supposed to be the instantaneous slope. It seems like they are separate; deriving one function is the opposite of integrating it but when we look at it graphically it doesn't make sense. Also, if I start with some function, differentiate it and then take the integral I get the original function so how is that the area under the curve? Thanks.
Calculus – Understanding Integration as the Opposite of Differentiation
calculus
Related Solutions
The magic word is "rate of change". The slope is the rate of change, so the slope is simply "how much the function grows when you move right". So if you plot the integral curve (which geometrically can be interpreted as an area under some other curve), then the rate of change is "how much the area grows when you expand it to the right", which is exactly the value of the original function.
Sketch:
On the left, you have a curve of the original function $f(x)$, and the approximate area under it, shaded as red rectangles. The curve of the integral $\int f(x)dx$ is obtained by "adding" the areas together (right figure, the $y$ coordinate measures the total area of the curve in the left figure). The derivative of the right curve is the slope (dashed lines across the rectangles), and obviously, the slope of a rectangle with unit width is exactly its height, and the height of the rectangle brings you back to the left figure and the value $f(x)$.
I think the first FTC:
If $f: [a,b] \to \Bbb R$ is continuous then $F: [a,b] \to \Bbb R$ defined by $F(x)=\int_a^x f(t)dt$ is differentiable and $F'(x)=f(x)$ for all $x \in [a,b]$.
is what people mean by saying the integration (which defines $F$) is the inverse of differentiation (as we have found a function with derivative $f$).
The second FTC
If $f: [a,b] \to \Bbb R$ is Riemann-integrable on $[a,b]$ and we have a function $F: [a,b] \to \Bbb R$ such that $F'(x)=f(x)$ on $[a,b]$, then $\int_a^b f(x)dx=F(b)-F(a)$.
is more of a "recipe" to find an integral: the target is to compute the definite integral and the tool we're given is to find an antiderivative. So not an inverse as such but a method. It's a bit of an iffy one, as an antiderivative $F$ need not exist at all (except when $f$ is continuous and the first FTC gives us one, but not explicitly, but at least we know some solution exists, but we don't have it in computable form yet). I think the first is closer to giving a direct "inverse" connection between integration and differentiation (and is often used in other contexts when we differentiate wrt boundaries of integrals, etc.). But that's just one view.
The first FTC can be summarised as $$\frac{d}{dx}\int_a^x f(t)dt = f(x)$$ so "Applying the integration operator to $f$, followed by the differentiation operator gives us back $f$ again".
Best Answer
Well...it's not obvious, as you point out, why "finding a slope" and "finding area under a curve" are opposites. That non-obviousness is why it's called a "theorem" (indeed, more formally stated, it's called "the fundamental theorem of calculus").
But maybe I can help out with the intuition a little. Instead of thinking of $f'(b)$ as the slope of the graph of $f$ at the point $(b, f(b))$, think of it this way: if we moved a little to the right of $b$, say, to $b + h$, where $h$ is a small number, what would we expect $f(b+h)$ to be? Well, if you draw a picture, you'll see that we expect it to be
$$ f(b+h) \approx f(b) + h * slope $$ and since the slope is $f'(b)$, this is
$$ f(b + h) \approx f(b) + h f'(b). $$
That's true for any "nice" (smooth, etc.) function, for small values of $h$. Let me define such a function, the "accumulated area" function:
$$ F(x) = \int_0^x f(t) ~dt. $$
That's the area under the graph of $f$ between $0$ and $x$. In particular, we have
$$ F(b) = \int_0^b f(t) ~dt $$ is the area under the graph of $f$ from $0$ to $b$. What's the value of $F(b + h)$ for a small value of $h$?
We have two possible answers: the first is from the definition: it's just
$$ F(b+h) = \int_0^{b+h} f(t) ~dt = \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt. $$
Hold that thought.
The second answer comes from that general stuff I wrote above, namely: $$ F(b+h) \approx F(b) + h F'(b). $$ Setting these two equal (since they both represent the value of $F(b+h)$, more or less), we get $$ F(b) + h F'(b) \approx \int_0^{b} f(t) ~dt + \int_b^{b+h} f(t) ~dt $$ Since the $F(b)$ on the left hand side is the same as the first integral on the right hand side, this simplifies to $$ h F'(b) \approx \int_b^{b+h} f(t) ~dt $$ and dividing through by $h$, becomes $$ F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt $$
Now when $h$ is very small, the integrand, $f(t)$ is roughly constant ... and its value is approximately $f(b)$, So the integral on the right is roughly the integral of the constant $f(b)$ over an interval of width $h$; that value is just $h f(b)$. So we get
$$ F'(b) \approx \frac{1}{h} \int_b^{b+h} f(t) ~dt \approx \frac{1}{h} [h f(b) ] = f(b). $$
So the derivative of the "accumulated area" function $F$ is the original function $f$.
As for derivative and integral being "opposites", you might want to look at
$$ f(x) = x^2 + 1. $$ Try taking its derivative, to get a new function $g$, and then write down the accumulated area function $$ G(x) = \int_0^x g(t) dt. $$ Evaluate the integral, and see whether it equals $f$. (Hint: it doesn't!). So "differentiate then integrate" doesn't necessarily bring you back to the original function. On the other hand, for nice enough functions (e.g., continuous), "integrate then differentiate" does bring you back to the original.