Derivatives of Integral Functions

Section 07.05 showed that $g(x) = \int_a^x f(t)\,dt$ has derivative $g'(x) = f(x)$. That means a graph of $f$ is a graph of $g$'s slope, and you can analyze $g$ without ever finding a formula for it.

What $f$ tells you about $g$

About $g$	Read it from $f$
value $g(x)$	signed area under $f$ from $a$ to $x$
$g$ increasing / decreasing	$f > 0$ / $f < 0$
local max / min of $g$	$f$ changes from $+$ to $-$ / from $-$ to $+$
$g$ concave up / down	$f$ increasing / decreasing (since $g'' = f'$)
$g(a)$	always $0$

Example

Here $f$ is made of straight pieces, so every area is a triangle or a rectangle. Let $g(x) = \int_0^x f(t)\,dt$.

• $g(1) = -1$ (a triangle below the axis), $g(2) = -1 + 1 = 0$, $g(4) = 0 + 4 = 4$, $g(5) = 5$, $g(6) = 4$.
• $g$ decreases on $(0, 1)$ and increases on $(1, 5)$, so $g$ has a local min at $x = 1$ and a local max at $x = 5$.
• $f$ is decreasing on $(4, 6)$, so $g$ is concave down there.

Function limits, both ends

With the chain rule (07.05):

$$\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt = f\big(u(x)\big)\,u'(x) - f\big(v(x)\big)\,v'(x)$$

Tangent lines

For $g(x) = \int_1^x \sqrt{t^3 + 3}\,dt$ at $x = 1$: $g(1) = 0$ and $g'(1) = \sqrt{4} = 2$, so the tangent line is $y = 2(x - 1)$. No antiderivative needed.

Limits

An accumulation function with matching limits starts at $0$, so limits involving one are often $\frac{0}{0}$. L'Hôpital's rule (06.10) and the theorem handle them together:

$$\lim_{x \to 0}\frac{\int_0^x \cos(t^2)\,dt}{x} = \lim_{x\to 0}\frac{\cos(x^2)}{1} = 1$$