Unit 6 Notes: Understanding the Fundamental Theorem of Calculus (AP Calculus BC)

FTC and Accumulation Functions

What an accumulation function is (and why you should care)

An accumulation function is a function built from an integral that “keeps track” of how much total change has built up from a starting point to a variable endpoint. The key idea is that definite integrals measure accumulated change (like total distance from velocity, or total mass from a density), and if you let the upper (or lower) limit be a variable, you get a new function whose output changes as that endpoint moves.

A standard accumulation function looks like this:

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

Here’s what each piece means:

• $f(t)$ is the rate or density you are accumulating.
• $a$ is the fixed starting point.
• $x$ is the variable endpoint.
• $g(x)$ is the accumulated net amount from $a$ to $x$.

This matters because it links two big ideas in calculus:

• Differentiation: instantaneous rate of change
• Integration: accumulated change

The Fundamental Theorem of Calculus (FTC) is exactly the bridge between them.

The Fundamental Theorem of Calculus, Part 1 (FTC1)

FTC1 tells you how to differentiate an accumulation function.

If $g(x) = \int_a^x f(t)\,dt$ and $f$ is continuous, then

$g'(x) = f(x)$

Why this is true (conceptually)

Think about increasing $x$ by a tiny amount $\Delta x$. The accumulation $g(x)$ increases by approximately the area of a thin rectangle:

• width $\Delta x$
• height approximately $f(x)$

So the change is approximately $f(x)\Delta x$, and dividing by $\Delta x$ gives a rate of change about $f(x)$. Taking the limit makes that exact.

Differentiating more complicated accumulation functions

AP Calculus often uses variations where the variable shows up in places other than “upper limit equals $x$.” The core strategy is: use FTC1 plus the Chain Rule, and pay attention to orientation.

Case 1: Upper limit is a function of $x$

$g(x) = \int_a^{h(x)} f(t)\,dt$

Then

$g'(x) = f(h(x))\cdot h'(x)$

You’re doing FTC1 (replace $x$ by $h(x)$), then multiplying by $h'(x)$ because the endpoint is moving at rate $h'(x)$.

Case 2: Lower limit is a function of $x$

$g(x) = \int_{h(x)}^a f(t)\,dt$

You can rewrite by reversing limits:

$\int_{h(x)}^a f(t)\,dt = -\int_a^{h(x)} f(t)\,dt$

So

$g'(x) = -f(h(x))\cdot h'(x)$

A common mistake is forgetting this negative sign when the variable is in the lower limit.

Case 3: Both limits depend on $x$

$g(x) = \int_{u(x)}^{v(x)} f(t)\,dt$

Rewrite as a difference:

$g(x) = \int_a^{v(x)} f(t)\,dt - \int_a^{u(x)} f(t)\,dt$

Differentiate:

$g'(x) = f(v(x))\cdot v'(x) - f(u(x))\cdot u'(x)$

Worked examples (FTC1)

Example 1: Basic accumulation derivative

Let

$g(x) = \int_2^x \sqrt{t^2+1}\,dt$

By FTC1,

$g'(x) = \sqrt{x^2+1}$

The dummy variable $t$ becomes $x$ after differentiating.

Example 2: Chain rule with an inside function

Let

$g(x) = \int_0^{x^3} \cos(t)\,dt$

Differentiate:

$g'(x) = \cos(x^3)\cdot 3x^2$

Example 3: Variable in the lower limit

Let

$g(x) = \int_x^5 \ln(t)\,dt$

Rewrite:

$g(x) = -\int_5^x \ln(t)\,dt$

Differentiate:

$g'(x) = -\ln(x)$

Notation reference (common FTC forms)

Exam Focus

• Typical question patterns:
  • Differentiate a function defined by an integral with variable limits, sometimes with a composite limit like $x^2+1$.
  • Find $g'(c)$ at a point using a table/graph of $f$ (you plug in, not integrate).
  • Write an expression for a new function as an accumulation function and then differentiate it.
• Common mistakes:
  • Forgetting the Chain Rule when the upper limit is $h(x)$.
  • Dropping the negative sign when the variable is in the lower limit.
  • Confusing the dummy variable (like $t$) with the outside variable (like $x$).

Interpreting the Behavior of Accumulation Functions

Turning “area under a curve” into function behavior

When you define

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

you can learn how $g$ behaves (increasing/decreasing, concavity, extrema) from the graph or sign of $f$—often without computing any integrals exactly.

The reason is FTC1 gives you an immediate relationship:

$g'(x) = f(x)$

So the derivative of the accumulation function is literally the original integrand.

Increasing and decreasing

A function increases where its derivative is positive. Since $g'(x)=f(x)$:

• $g$ is **increasing** where $f(x) > 0$.
• $g$ is **decreasing** where $f(x) < 0$.

Interpretation-wise: if the “rate” $f$ is positive, the accumulation is growing; if the “rate” is negative, the accumulation is shrinking.

Example 1: Monotonicity from a sign chart

Suppose a graph shows $f(x)$ is positive on $1<x<4$ and negative on $4<x<6$. If

$g(x)=\int_1^x f(t)\,dt$

then $g$ increases on $1<x<4$ and decreases on $4<x<6$.

A common misconception is to say “$g$ is positive where $f$ is positive.” That’s not necessarily true—$g(x)$ depends on total accumulated area from $a$ to $x$, including what happened earlier.

Where does an accumulation function have a maximum or minimum?

Local extrema occur where $g'(x)=0$ and the sign of $g'$ changes. Since $g'(x)=f(x)$, extrema of $g$ happen when:

$f(x)=0$

and $f$ changes sign.

• If $f$ changes from positive to negative, $g$ has a local maximum.
• If $f$ changes from negative to positive, $g$ has a local minimum.

This is a powerful AP move: you can find maxima/minima of $g$ from the graph of $f$.

Concavity of an accumulation function

Concavity depends on the second derivative. Differentiate again:

$g'(x)=f(x)$

So

$g''(x)=f'(x)$

That means:

• $g$ is **concave up** where $f'(x)>0$ (where $f$ is increasing).
• $g$ is **concave down** where $f'(x)<0$ (where $f$ is decreasing).

This often surprises students: concavity of $g$ depends on whether $f$ is increasing or decreasing, not whether $f$ is positive or negative.

Interpreting values of $g(x)$ as signed area

By definition,

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

is the net signed area between $f$ and the horizontal axis from $t=a$ to $t=x$.

• Areas above the axis contribute positively.
• Areas below the axis contribute negatively.

So $g(x)$ could be negative even if $x>a$, as long as the area below the axis dominates.

“Accumulation as total change” (real-world interpretation)

If $f(t)$ is a rate of change (like gallons per minute flowing into a tank, or velocity in meters per second), then

$\int_a^x f(t)\,dt$

gives net change in the quantity from time $a$ to time $x$.

• If $f(t)$ is velocity, the integral is displacement (not total distance unless velocity stays nonnegative).
• If $f(t)$ is a “net rate in,” negative values represent net outflow.

Worked example: Sketching an accumulation function from a graph

Suppose $f$ is given by a graph with these features:

• $f(x)>0$ for $0<x<2$
• $f(x)=0$ at $x=2$
• $f(x)<0$ for $2<x<5$
• $f$ is increasing on $0<x<1$ and decreasing on $1<x<5$

Let

$g(x)=\int_0^x f(t)\,dt$

Then:

1. Increasing/decreasing: $g$ increases on $0<x<2$ and decreases on $2<x<5$.
2. Extrema: $g$ has a local maximum at $x=2$ (since $f$ changes from positive to negative).
3. Concavity: $g$ is concave up on $0<x<1$ (since $f$ increasing) and concave down on $1<x<5$ (since $f$ decreasing).

Notice you didn’t need exact areas to get the overall shape.

Exam Focus

• Typical question patterns:
  • Given a graph of $f$, analyze where $g(x)=\int_a^x f(t)dt$ is increasing/decreasing, and where it has relative extrema.
  • Given a table of values for $f$, estimate or compare values of $g$ at different points using areas (often trapezoids or rectangles).
  • Use $g''(x)=f'(x)$ logic: determine concavity of $g$ from whether $f$ is increasing/decreasing.
• Common mistakes:
  • Treating $g(x)$ as “the same as $f(x)$” instead of accumulated area.
  • Confusing “$f(x)>0$” with “$g(x)>0$.” Sign of the rate is not the same as sign of total accumulated change.
  • Mixing up concavity: concavity of $g$ depends on $f'$, not on the sign of $f$.

Applying Properties of Definite Integrals

Why properties matter

You do not always need to compute an antiderivative to work with definite integrals. The properties of definite integrals let you simplify expressions, evaluate integrals using known values, and reason about signs and sizes of areas. On AP exams, these properties show up when you are given a few integral values (or a graph) and asked to find a new integral quickly.

In all properties below, assume the integrals exist (for AP-level problems, this is usually ensured by continuity on the interval).

Linearity (splitting sums and constants)

Definite integrals behave like “area measurements,” so they distribute over addition and allow constants to factor out.

If $c$ is a constant, then:

$\int_a^b \left(f(x)+g(x)\right)\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$

$\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx$

This is especially useful when an integrand is a sum of parts you already know.

Reversing limits changes the sign

Changing direction flips the sign:

$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

Also,

$\int_a^a f(x)\,dx = 0$

A common error is to reverse limits without adding the negative sign.

Additivity across intervals

If you break an interval at a point $c$, the total integral is the sum:

$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$

This is the “accumulation adds” idea: net change from $a$ to $b$ equals net change from $a$ to $c$ plus net change from $c$ to $b$.

Symmetry: even and odd functions (when limits are symmetric)

These are high-leverage shortcuts when you integrate over $[-a,a]$.

• If $f$ is **even** (meaning $f(-x)=f(x)$), then

$\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$

• If $f$ is **odd** (meaning $f(-x)=-f(x)$), then

$\int_{-a}^a f(x)\,dx = 0$

This is about cancellation: odd functions have equal area above and below the axis on symmetric intervals.

Comparing integrals using inequalities

If $f(x)\ge g(x)$ for all $x$ in $[a,b]$, then

$\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx$

A particularly common special case: if $f(x)\ge 0$ on $[a,b]$, then

$\int_a^b f(x)\,dx \ge 0$

This helps you reason about signs without computation.

Worked examples with properties

Example 1: Using given integral values

Suppose you know

$\int_0^5 f(x)\,dx = 12$

and

$\int_0^5 g(x)\,dx = -3$

Find

$\int_0^5 \left(2f(x)-5g(x)\right)\,dx$

Use linearity:

$\int_0^5 \left(2f(x)-5g(x)\right)\,dx = 2\int_0^5 f(x)\,dx - 5\int_0^5 g(x)\,dx$

Substitute values:

$= 2(12) - 5(-3) = 24 + 15 = 39$

Example 2: Interval additivity and reversing limits

Suppose

$\int_1^4 f(x)\,dx = 7$

and

$\int_4^6 f(x)\,dx = -2$

Then

$\int_1^6 f(x)\,dx = \int_1^4 f(x)\,dx + \int_4^6 f(x)\,dx = 7 + (-2) = 5$

Also,

$\int_6^1 f(x)\,dx = -\int_1^6 f(x)\,dx = -5$

Example 3: Odd symmetry shortcut

If $f$ is odd, then

$\int_{-3}^3 f(x)\,dx = 0$

Even if you have no formula for $f$, this conclusion is immediate from symmetry.

Exam Focus

• Typical question patterns:
  • You are given a few definite integral values (like $\int_0^2 f$ and $\int_2^5 f$) and asked to compute a new one using properties.
  • Simplify expressions like $\int_a^b (3f(x)+2)dx$ without finding antiderivatives.
  • Use symmetry (even/odd) on integrals over $[-a,a]$.
• Common mistakes:
  • Forgetting that reversing limits introduces a negative.
  • Breaking an interval incorrectly (mixing up endpoints or signs).
  • Misidentifying even vs odd: even means symmetric about the $y$-axis; odd means origin symmetry.

FTC and Evaluating Definite Integrals

The Fundamental Theorem of Calculus, Part 2 (FTC2)

FTC2 is the “evaluation engine” for definite integrals. It says that if you can find an antiderivative, then a definite integral becomes a subtraction problem.

If $F'(x)=f(x)$ on $[a,b]$, then

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

This result is often written using the notation:

$\int_a^b f(x)\,dx = \left[F(x)\right]_a^b$

where

$\left[F(x)\right]_a^b = F(b)-F(a)$

Why FTC2 makes sense

Integration (as a definite integral) measures net accumulation. Antiderivatives measure the “original quantity” whose derivative is the rate. FTC2 formalizes the idea that:

• adding up the rate over an interval gives total change,
• and an antiderivative tracks that change.

So the net accumulated change from $a$ to $b$ is the difference between the antiderivative at the endpoints.

The Net Change Theorem (a key interpretation)

A very common AP interpretation is:

If $R(t)$ is the rate of change of a quantity $Q(t)$, so that

$Q'(t)=R(t)$

then

$\int_a^b R(t)\,dt = Q(b)-Q(a)$

This is FTC2 in context. It’s how you justify statements like:

• Integral of velocity equals displacement.
• Integral of a growth rate equals total growth.

Worked examples (FTC2)

Example 1: Evaluate a definite integral using an antiderivative

Evaluate

$\int_1^3 \left(2x^3-4x\right)\,dx$

Step 1: Find an antiderivative $F(x)$.

$\int \left(2x^3-4x\right)dx = \frac{2}{4}x^4 -2x^2 = \frac{1}{2}x^4 - 2x^2$

So

$F(x)=\frac{1}{2}x^4-2x^2$

Step 2: Apply FTC2.

$\int_1^3 \left(2x^3-4x\right)dx = F(3)-F(1)$

Compute:

$F(3)=\frac{1}{2}(81)-2(9)=\frac{81}{2}-18=\frac{81}{2}-\frac{36}{2}=\frac{45}{2}$

$F(1)=\frac{1}{2}(1)-2(1)=\frac{1}{2}-2=\frac{1}{2}-\frac{4}{2}=-\frac{3}{2}$

Subtract:

$F(3)-F(1)=\frac{45}{2}-\left(-\frac{3}{2}\right)=\frac{48}{2}=24$

Example 2: Connecting to a real context (net change)

A tank’s water volume $V(t)$ changes at rate $V'(t)=r(t)$ (in liters per minute). Suppose

$r(t)=5-2t$

for $0\le t\le 3$. If $V(0)=100$, find $V(3)$.

First compute the net change:

$\int_0^3 (5-2t)\,dt$

Antiderivative:

$\int (5-2t)dt = 5t-t^2$

Evaluate:

$\int_0^3 (5-2t)dt = (5(3)-3^2)-(5(0)-0^2) = (15-9)-0=6$

So the volume increased by 6 liters:

$V(3)=V(0)+6=106$

A typical misconception here is to treat negative values of $r(t)$ as “impossible.” In many contexts negative rate just means the quantity is decreasing (water draining faster than it flows in, for example).

Using FTC1 together with FTC2 (a powerful combo)

Sometimes you define

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

and later need a definite integral like $\int_p^q f(x)dx$. FTC2 can be applied using $G$ itself, because $G'(x)=f(x)$, meaning $G$ is an antiderivative of $f$.

So,

$\int_p^q f(x)\,dx = G(q)-G(p)$

This is especially useful if $G$ is given (maybe as a graph or table), even when $f$ is not easy to integrate algebraically.

Example 3: Evaluating an integral using an accumulation function

Let

$G(x)=\int_2^x f(t)\,dt$

Suppose you are told $G(10)=7$ and $G(4)=-1$. Find

$\int_4^{10} f(x)\,dx$

Because $G$ is an antiderivative of $f$,

$\int_4^{10} f(x)dx = G(10)-G(4)=7-(-1)=8$

No antiderivative formula for $f$ was needed.

What can go wrong when evaluating definite integrals

1. Forgetting the “plus C” is unnecessary: For definite integrals, constants cancel in $F(b)-F(a)$, so you do not include a constant of integration.
2. Sign errors with bounds: Most wrong answers come from evaluating $F(a)-F(b)$ instead of $F(b)-F(a)$.
3. Confusing net area with total area: FTC2 gives net signed area. If a problem asks for total distance (not displacement), you must handle where velocity changes sign by splitting the interval.

Exam Focus

• Typical question patterns:
  • Evaluate a definite integral by finding an antiderivative and applying $F(b)-F(a)$.
  • Use the Net Change Theorem: given a rate function and an initial value, find the final value.
  • Given an accumulation function $G(x)=\int_a^x f(t)dt$, compute definite integrals like $\int_p^q f(x)dx$ using $G(q)-G(p)$.
• Common mistakes:
  • Writing $F(a)-F(b)$ instead of $F(b)-F(a)$.
  • Adding a constant of integration when evaluating a definite integral.
  • Treating the integral as total area when the problem is actually about net change (or vice versa).