Fundamental Theorem
of Calculus

Master the deep connection between differentiation and integration through accumulation functions and their derivatives.

AP Calculus BC

What is an Accumulation Function?

An accumulation function measures the total "accumulated" quantity of a function from a fixed starting point $a$ up to a variable endpoint $x$. It is defined as:

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

As $x$ changes, the integral accumulates more (or less) signed area under the curve $f(t)$. The value $F(x)$ is the net signed area between $t = a$ and $t = x$.

Follow the salt through a working mixing rig

A stirred transparent tank has one upper inlet and one lower outlet with a valve. The exact diagram links its 100 L salt balance to a computed approach to equilibrium.

Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

Net inventory versus throughput

Suppose a tank receives and releases 4 gal/min for 5 minutes. Its 200 gal inventory stays constant while 20 gal enters and 20 gal leaves: net change is zero and total throughput is 40 gal.

Visualizing Signed Area

Why Does This Matter?

Physics

Velocity integrated over time gives displacement: position is the accumulation of velocity.

Biology

A population growth rate integrated over time gives the total change in population size.

Economics

Marginal cost integrated over units produced gives total variable cost of production.

FTC Part 1: Differentiation of Integrals

If $f$ is continuous on $[a, b]$ and we define:

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

then $F$ is differentiable on $(a, b)$ and:

$$ \boxed{\frac{d}{dx}\int_a^x f(t)\, dt = f(x)} $$

Seeing FTC Part 1 Graphically

FTC Part 2: Evaluation Theorem

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F'(x) = f(x)$), then:

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

Part 1 vs Part 2

	FTC Part 1	FTC Part 2
Statement	$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$	$\int_a^b f(x)\,dx = F(b) - F(a)$
Direction	Integral → Derivative	Derivative → Integral
Purpose	Differentiate an accumulation function	Evaluate a definite integral
Output	A function: $f(x)$	A number: $F(b) - F(a)$
Requirement	$f$ continuous on $[a, b]$	$F' = f$ on $[a, b]$

FTC Part 1 with the Chain Rule

When the upper limit is not simply $x$ but a function $g(x)$, we apply the chain rule:

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

Example: $\frac{d}{dx}\int_1^{x^2} \cos(t)\,dt = \cos(x^2) \cdot 2x$.

Variable Lower and Upper Limits

What if both limits depend on $x$? Split using the additive property of integrals:

$$ \frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\, dt = f\!\big(g(x)\big)\cdot g'(x) \;-\; f\!\big(h(x)\big)\cdot h'(x) $$

1

Split at a constant $c$: $\int_{h(x)}^{g(x)} = \int_c^{g(x)} - \int_c^{h(x)}$

2

Differentiate each part using FTC Part 1 + chain rule

3

Combine: Upper limit contributes $+f(g(x))\cdot g'(x)$; lower contributes $-f(h(x))\cdot h'(x)$

How to Apply FTC Part 1

1

Identify the integrand $f(t)$ and the upper limit function $g(x)$

2

Substitute $t = g(x)$ into the integrand to get $f(g(x))$

3

Multiply by $g'(x)$ (the chain rule factor)

4

Check the lower limit: if it also depends on $x$, subtract its analogous term

5

Simplify the resulting expression

Example 1: Basic FTC Part 1

Find $\displaystyle\frac{d}{dx}\int_2^x (3t^2 + 1)\, dt$

1

Integrand: $f(t) = 3t^2 + 1$. Upper limit: $g(x) = x$

2

Substitute $t = x$: $f(x) = 3x^2 + 1$

3

Chain rule factor: $g'(x) = 1$

Example 2: Chain Rule Required

Find $\displaystyle\frac{d}{dx}\int_0^{x^3} e^{-t^2}\, dt$

1

Integrand: $f(t) = e^{-t^2}$. Upper limit: $g(x) = x^3$

2

Substitute $t = x^3$: $f(x^3) = e^{-(x^3)^2} = e^{-x^6}$

3

Chain rule factor: $g'(x) = 3x^2$

Example 3: Both Limits Variable

Find $\displaystyle\frac{d}{dx}\int_{\sin x}^{x^2} \ln(t+1)\, dt$

1

Upper limit: $g(x) = x^2$, so $g'(x) = 2x$. Contribution: $\ln(x^2 + 1) \cdot 2x$

2

Lower limit: $h(x) = \sin x$, so $h'(x) = \cos x$. Contribution: $-\ln(\sin x + 1) \cdot \cos x$

3

Combine the terms

Properties of Accumulation Functions

$F(a) = 0$

The accumulation starts at zero since $\int_a^a f(t)\,dt = 0$

Increasing

$F$ is increasing where $f(x) > 0$ (positive integrand adds area)

Decreasing

$F$ is decreasing where $f(x) < 0$ (negative integrand subtracts area)

Extrema

$F$ has critical points where $f(x) = 0$ (sign change in $f$ determines max/min)

Common Mistakes to Avoid

✕ Forgetting the Chain Rule

When the upper limit is $g(x) \neq x$, you must multiply by $g'(x)$. Writing $f(g(x))$ alone is incomplete.

✕ Wrong Variable Substitution

Replace the dummy variable $t$ with the upper limit function, not with $x$ blindly. $f(t)$ becomes $f(g(x))$, not $f(x)$.

✕ Sign Error on Lower Limit

A variable lower limit contributes a negative term. Forgetting the minus sign is a common error on AP exams.

Question 1 of 5

What is $\displaystyle\frac{d}{dx}\int_1^x \sqrt{t^3 + 4}\, dt$?

By FTC Part 1, $\frac{d}{dx}\int_1^x f(t)\,dt = f(x)$. Here the upper limit is simply $x$, so we substitute $t = x$ into $\sqrt{t^3+4}$ to get $\sqrt{x^3+4}$. No chain rule factor is needed since $g(x) = x$ and $g'(x)=1$. Choice (c) incorrectly applies the chain rule as if we were differentiating the integrand.

Question 2 of 5

What is $\displaystyle\frac{d}{dx}\int_0^{\sin x} t^2\, dt$?

The upper limit is $g(x) = \sin x$, so we need the chain rule. Substitute $t = \sin x$ into $t^2$ to get $\sin^2(x)$, then multiply by $g'(x) = \cos x$. The answer is $\sin^2(x) \cdot \cos x$. Choice (a) forgets the chain rule factor.

Question 3 of 5

If $F(x) = \displaystyle\int_2^x f(t)\,dt$ and $f$ is continuous, which statement is always true?

Since $F(2) = \int_2^2 f(t)\,dt = 0$, the accumulation always starts at zero when evaluated at the lower bound. Choices (b), (c), and (d) are not guaranteed: if $f(t) < 0$ on some interval past $t = 2$, then $F$ would decrease and could become negative.

Question 4 of 5

What is $\displaystyle\frac{d}{dx}\int_{x}^{x^2} e^t\, dt$?

Both limits are variable. The upper limit $g(x) = x^2$ contributes $+e^{x^2}\cdot 2x$. The lower limit $h(x) = x$ contributes $-e^x \cdot 1$. Combining: $2x \cdot e^{x^2} - e^x$. Choice (a) forgets the chain rule on the upper limit. Choice (c) has the wrong sign on the lower limit term.

Question 5 of 5

Suppose $f$ is continuous, $f(3) = 5$, and $G(x) = \displaystyle\int_0^x f(t)\,dt$. What is the equation of the tangent line to $G$ at $x = 3$, given $G(3) = 7$?

By FTC Part 1, $G'(x) = f(x)$, so $G'(3) = f(3) = 5$. The tangent line at $x = 3$ uses point-slope form: $y - G(3) = G'(3)(x - 3)$, which gives $y = 5(x - 3) + 7$. Choice (a) swaps the slope and the y-value.

	FTC Part 1	FTC Part 2
Statement	\(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\)	\(\int_a^b f(x)\,dx = F(b) - F(a)\)
Direction	Integral → Derivative	Derivative → Integral
Purpose	Differentiate an accumulation function	Evaluate a definite integral
Output	A function: \(f(x)\)	A number: \(F(b) - F(a)\)
Requirement	\(f\) continuous on \([a, b]\)	\(F' = f\) on \([a, b]\)

Fundamental Theoremof Calculus

What is an Accumulation Function?

Follow the salt through a working mixing rig

Net inventory versus throughput

Visualizing Signed Area

Why Does This Matter?

FTC Part 1: Differentiation of Integrals

Seeing FTC Part 1 Graphically

FTC Part 2: Evaluation Theorem

Part 1 vs Part 2

FTC Part 1 with the Chain Rule

Variable Lower and Upper Limits

How to Apply FTC Part 1

Example 1: Basic FTC Part 1

Example 2: Chain Rule Required

Example 3: Both Limits Variable

Properties of Accumulation Functions

Common Mistakes to Avoid

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Fundamental Theorem
of Calculus