Concept preview

What you will be able to do

1. 01State the Fundamental Theorem of Calculus in accumulation form: if $g(x)=\displaystyle\int_a^x f(t)\,dt$ then $g'(x)=f(x)$, and interpret $g$ as the running total of area accumulated by $f$ from $a$ up to $x$.
2. 02Evaluate an accumulation function $g$ and its derivative $g'$ directly from a graph of $f$ — reading $g$ as signed area so far and $g'(x)$ as the height of $f$ at $x$.
3. 03Apply the chain-rule version $\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt=f\bigl(u(x)\bigr)\,u'(x)$ when the upper limit is itself a function of $x$.

Fix a starting point $a$ and let the upper limit vary. Define $g(x)=\int_a^x f(t)\,dt.$ For each input $x$, the output $g(x)$ is the signed area under the graph of $f$ from $t=a$ to $t=x$ — the total accumulated so far. We use $t$ as the dummy variable inside the integral to keep it distinct from the limit $x$; the letter $t$ is swept over as we integrate, while $x$ marks where the sweeping stops. When $x=a$ the region has zero width, so $g(a)=\int_a^a f\,dt=0$: accumulation always starts from nothing. As $x$ increases past $a$, $g$ grows by adding area where $f>0$ and shrinks by subtracting area where $f<0$ (signed area).