Accumulation Functions and the FTC

What if the upper limit of an integral is a variable? Then the integral becomes a function: it keeps a running total of the signed area as you move to the right. These accumulation functions lead to one of the most important results in all of mathematics, the Fundamental Theorem of Calculus, which says that integrating and differentiating undo each other.

Key ideas

Accumulation functions

Fix a starting point $a$ and define

$$g(x) = \int_a^x f(t)\,dt .$$

$g(x)$ is the signed area under $f$ from $a$ to $x$. The letter $t$ is just a placeholder (a “dummy variable”), so it doesn’t clash with $x$, which is now a limit. Two facts follow right away:

• $g(a) = \displaystyle\int_a^a f(t)\,dt = 0$.
• If $x \lt a$, the integral runs backwards, so $g(x) = -\displaystyle\int_x^a f(t)\,dt$.

The Fundamental Theorem of Calculus

If $f$ is continuous on an interval containing $a$, then $g(x) = \displaystyle\int_a^x f(t)\,dt$ is differentiable and

$$g'(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x).$$

Why it works: when $x$ moves a tiny bit, $h$, to the right, $g$ gains a thin strip of area that is about $f(x) \cdot h$. So the rate at which area accumulates is $\dfrac{f(x)\,h}{h} = f(x)$, the height of the graph.

The starting point $a$ doesn’t matter for the derivative: changing it only adds a constant to $g$.

Variable limits and the chain rule

If the upper limit is a function $u(x)$, use the chain rule:

$$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\big(u(x)\big)\cdot u'(x).$$

If the variable is in the lower limit, flip the limits first, which adds a negative sign:

$$\frac{d}{dx}\int_x^b f(t)\,dt = \frac{d}{dx}\left(-\int_b^x f(t)\,dt\right) = -f(x).$$

Reading g from the graph of f

Since $g' = f$ and $g'' = f'$, the graph of $f$ tells you everything about $g$:

On the graph of $f$	For $g(x) = \int_a^x f(t)\,dt$
$f$ above the axis ($f \gt 0$)	$g$ is increasing
$f$ below the axis ($f \lt 0$)	$g$ is decreasing
$f$ changes from positive to negative	$g$ has a relative maximum
$f$ changes from negative to positive	$g$ has a relative minimum
$f$ increasing	$g$ is concave up
$f$ decreasing	$g$ is concave down
$f$ changes from increasing to decreasing (or back)	$g$ has a point of inflection

Values of $g$ come from signed areas, often using geometry. For absolute extrema, compare $g$ at the critical points and at the endpoints (the Candidates Test).

Worked examples

Example 1: Using the FTC directly

Find each derivative.

(a) $\dfrac{d}{dx}\displaystyle\int_1^x \cos(t^2)\,dt$

(b) $\dfrac{d}{dx}\displaystyle\int_x^5 \sqrt{1 + t^3}\,dt$

Solution.

(a) The integrand is continuous, so by the FTC the derivative is the integrand with $t$ replaced by $x$: $\cos(x^2)$.

(b) The variable is the lower limit. Flip the limits:

$$\frac{d}{dx}\int_x^5 \sqrt{1 + t^3}\,dt = \frac{d}{dx}\left(-\int_5^x \sqrt{1 + t^3}\,dt\right) = -\sqrt{1 + x^3}$$

You don’t need to (and can’t easily) find either integral. The FTC goes straight to the derivative.

Example 2: A variable upper limit

Find $\dfrac{d}{dx}\displaystyle\int_2^{x^2} \ln(1 + t)\,dt$.

Solution. The upper limit is $u(x) = x^2$, with $u'(x) = 2x$. Put $u$ into the integrand and multiply by $u'$:

$$\frac{d}{dx}\int_2^{x^2} \ln(1 + t)\,dt = \ln(1 + x^2) \cdot 2x = 2x\ln(1 + x^2)$$

Example 3: Reading g from a graph

The graph of $f$ on $[0, 8]$ is shown. It is made of a line segment, a semicircle of radius $2$, and another line segment. Let $g(x) = \displaystyle\int_0^x f(t)\,dt$.

Signed areas under $f$: positive above the axis, negative below.

(a) Find $g(2)$, $g(4)$, $g(6)$, and $g(8)$.

(b) Find $g'(5)$ and $g''(1)$.

(c) On what intervals is $g$ increasing? Where does $g$ have relative extrema? Justify.

(d) Where is $g$ concave up? Where does it have a point of inflection?

(e) Find the absolute maximum and minimum values of $g$ on $[0, 8]$.

Solution.

(a) Add signed areas from $0$:

$$\begin{aligned} g(2) &= 2 && \text{triangle} \\ g(4) &= 2 - \pi && \text{minus a quarter circle, } \tfrac{1}{4}\pi(2)^2 = \pi \\ g(6) &= 2 - 2\pi && \text{minus the whole half circle} \\ g(8) &= 2 - 2\pi + 2 = 4 - 2\pi && \text{plus the second triangle} \end{aligned}$$

(b) $g'(5) = f(5)$. On the semicircle, $f(t) = -\sqrt{4 - (t - 4)^2}$, so $f(5) = -\sqrt{3}$. And $g''(1) = f'(1)$, the slope of the first segment, which is $\dfrac{0 - 2}{2 - 0} = -1$.

(c) $g' = f$, which is positive on $(0, 2)$ and $(6, 8)$ and negative on $(2, 6)$. So $g$ is increasing on $(0, 2)$ and $(6, 8)$. $g$ has a relative maximum at $x = 2$ because $g' = f$ changes from positive to negative there, and a relative minimum at $x = 6$ because $g' = f$ changes from negative to positive there.

(d) $g'' = f'$. $f$ is decreasing on $(0, 4)$ and increasing on $(4, 8)$, so $g$ is concave down on $(0, 4)$ and concave up on $(4, 8)$. There is a point of inflection at $x = 4$, where $g' = f$ changes from decreasing to increasing.

(e) Candidates: the endpoints and the critical points $x = 2$ and $x = 6$.

$x$	$0$	$2$	$6$	$8$
$g(x)$	$0$	$2$	$2 - 2\pi \approx -4.283$	$4 - 2\pi \approx -2.283$

The absolute maximum is $2$ (at $x = 2$) and the absolute minimum is $2 - 2\pi$ (at $x = 6$).

Example 4: A tangent line

Let $h(x) = 3 + \displaystyle\int_1^x e^{t^2 - 1}\,dt$. Find the equation of the tangent line to $h$ at $x = 1$.

Solution. $h(1) = 3 + \displaystyle\int_1^1 e^{t^2 - 1}\,dt = 3 + 0 = 3$.

$h'(x) = e^{x^2 - 1}$ by the FTC, so $h'(1) = e^0 = 1$.

$$y = 3 + 1(x - 1), \quad \text{or} \quad y = x + 2$$

Common mistakes

Forgetting the chain rule. $\dfrac{d}{dx}\displaystyle\int_0^{3x} e^{-t^2}\,dt$ is $3e^{-9x^2}$, not $e^{-9x^2}$. Whenever the limit is anything other than plain $x$, multiply by its derivative.

Missing the sign when $x$ is the lower limit. Flip the limits first and pick up a negative sign.

Mixing up f and g in graph questions. If the graph shows $f$ and $g(x) = \int_a^x f(t)\,dt$, then $g$ increases where $f$ is positive, not where $f$ is increasing. Where $f$ increases, $g$ is concave up.

Using “g is positive” when you mean “g is increasing”. $g(x)$ can be negative while increasing (like $g$ on $(6, 8)$ in Example 3). For increasing or decreasing, look at the sign of $f$; for the value of $g$, add the areas.

Weak justifications. On the AP exam, ”$g$ has a relative maximum at $x = 2$” earns credit only with a reason like “because $g'(x) = f(x)$ changes from positive to negative at $x = 2$”. Saying “the graph goes down” or "$f(2) = 0$" isn’t enough.

Forgetting g(a) = 0. The accumulation starts at the lower limit, so the value there is $0$ (plus any constant added outside the integral, like the $3$ in Example 4).

Practice

1. (Warm-up) Find $\dfrac{d}{dx}\displaystyle\int_0^x (t^3 + 1)^5\,dt$.

Solution

By the FTC, the derivative is $(x^3 + 1)^5$.

2. (Warm-up) Let $F(x) = \displaystyle\int_x^3 \sqrt{t + 1}\,dt$. Find $F'(x)$ and $F(3)$.

Solution

$F(x) = -\displaystyle\int_3^x \sqrt{t + 1}\,dt$, so $F'(x) = -\sqrt{x + 1}$.

$F(3) = \displaystyle\int_3^3 \sqrt{t + 1}\,dt = 0$.

3. (Warm-up) For $g$ in Example 3, find $g(1)$ and $g'(7)$.

Solution

From $0$ to $1$, the region under $f$ is a trapezoid with parallel sides $f(0) = 2$ and $f(1) = 1$ and width $1$:

$$g(1) = \frac{2 + 1}{2} \cdot 1 = 1.5$$

$g'(7) = f(7) = 1$ (the last segment goes from $(6, 0)$ to $(8, 2)$).

4. (Core) Find $\dfrac{d}{dx}\displaystyle\int_0^{3x} e^{-t^2}\,dt$.

Solution

With $u = 3x$ and $u' = 3$:

$$e^{-(3x)^2} \cdot 3 = 3e^{-9x^2}$$

5. (Core) Find $\dfrac{d}{dx}\displaystyle\int_x^{x^2} \sin t\,dt$. (Hint: split the integral at $0$.)

Solution

$$\int_x^{x^2} \sin t\,dt = \int_0^{x^2} \sin t\,dt - \int_0^{x} \sin t\,dt$$

Differentiate each piece (chain rule on the first):

$$\frac{d}{dx}\int_x^{x^2} \sin t\,dt = \sin(x^2)\cdot 2x - \sin x = 2x\sin(x^2) - \sin x$$

6. (Core) Let $G(x) = 4 + \displaystyle\int_2^x \sqrt{t^2 + 5}\,dt$. Write the equation of the tangent line to $G$ at $x = 2$.

Solution

$G(2) = 4 + 0 = 4$. $G'(x) = \sqrt{x^2 + 5}$, so $G'(2) = \sqrt{9} = 3$.

$$y = 4 + 3(x - 2)$$

7. (Core) The graph of $f$ is shown in Example 3, and $g(x) = \displaystyle\int_0^x f(t)\,dt$. Is $g$ increasing or decreasing at $x = 3$? Is it concave up or concave down at $x = 3$? Give a reason for each.

Solution

$g'(3) = f(3) \lt 0$ (the semicircle is below the axis), so $g$ is decreasing at $x = 3$.

$g''(3) = f'(3) \lt 0$ because $f$ is decreasing at $x = 3$ (the semicircle is still heading down toward its lowest point at $x = 4$). So $g$ is concave down at $x = 3$.

8. (Challenge) Using the graph of $f$ in Example 3, let $h(x) = \displaystyle\int_2^x f(t)\,dt$. Find $h(0)$ and $h(8)$.

Solution

$$h(0) = \int_2^0 f(t)\,dt = -\int_0^2 f(t)\,dt = -2$$$$h(8) = \int_2^6 f(t)\,dt + \int_6^8 f(t)\,dt = -2\pi + 2$$

So $h(0) = -2$ and $h(8) = 2 - 2\pi \approx -4.283$. (Notice $h(x) = g(x) - 2$: changing the starting point shifts the function down.)

9. (Challenge) Let $g(x) = \displaystyle\int_0^x (t^2 - 4t + 3)\,dt$. Find the $x$-values where $g$ has relative extrema and points of inflection, and justify each.

Solution

By the FTC, $g'(x) = x^2 - 4x + 3 = (x - 1)(x - 3)$.

$g'$ is positive for $x \lt 1$, negative for $1 \lt x \lt 3$, and positive for $x \gt 3$.

• Relative maximum at $x = 1$, because $g'$ changes from positive to negative there.
• Relative minimum at $x = 3$, because $g'$ changes from negative to positive there.

$g''(x) = 2x - 4$, which changes from negative to positive at $x = 2$. So $g$ has a point of inflection at $x = 2$.