Single-Variable-Calculus · Unit 16 · Video 2 · Interactive Practice

Second Fundamental Theorem: Differentiate an Integral

IKey Formulas

Statement	Formula	Condition
FTC 2	$G(x) = \int_a^x f(t)\,dt \;\Longrightarrow\; G'(x) = f(x)$	$f$ continuous
Starting value	$G(a) = \int_a^a f(t)\,dt = 0$	an interval of length zero has no area
Differential equation	$y' = f(x),\ y(a) = 0$ is solved by $y = G(x)$	always solvable; Riemann sums compute it
Example	$\dfrac{d}{dx}\displaystyle\int_1^x \frac{dt}{t^2} = \frac{1}{x^2}$	recopy the integrand with $x$ in place of $t$

Key Insight: Integrate, then differentiate, and you are back at $f$ . The upper limit $x$ and the dummy variable $t$ do different jobs, and no $+c$ survives: changing the lower limit shifts $G$ by a constant, which differentiation erases.

IIVisualization 1 — The Area Grows at the Rate $f(x)$

$G(x) = \int_1^x \frac{dt}{t^2}$ is the area from $1$ to $x$ ; its slope at $x$ is the height $f(x)$ .

Upper limit x 2

IIIVisualization 2 — No $+c$ : Moving the Lower Limit

Changing the lower limit $a$ shifts $G_a(x) = \int_a^x \frac{dt}{t^2}$ up or down by a constant, so $G_a'$ never changes.

Lower limit a 1

IVQuiz Questions

Problem 1 · Recopy the Integrand

Find $\dfrac{d}{dx}\displaystyle\int_0^x \cos(t^2)\,dt$ .

✅ Correct! By FTC 2 the derivative is the integrand evaluated at the upper limit:

\cos(x^2)

. No antiderivative of

\cos(t^2)

is needed.

❌ That differentiates the integrand. The derivative of the integral is the integrand itself,

f(x) = \cos(x^2)

, not

f'(x)

❌ The lower limit contributes no term.

G(x) = \int_0^x f(t)\,dt

has

G'(x) = f(x)

; the value

f(0) = 1

never enters.

❌ Not quite. Replace

t

x

in the integrand.

Show solution

The function has the form $G(x) = \int_a^x f(t)\,dt$ with $a = 0$ and $f(t) = \cos(t^2)$ , which is continuous. FTC 2 gives

\frac{d}{dx}\int_0^x \cos(t^2)\,dt = \cos(x^2).

Problem 2 · Value and Slope at the Lower Limit

Given: $G(x) = \displaystyle\int_4^x \sqrt{t}\,dt$ . Find $G(4)$ and $G'(4)$ .

✅ Correct! The interval

[4, 4]

has length zero, so

G(4) = 0

; and

G'(4) = \sqrt{4} = 2

❌ $\tfrac{16}{3}$ is the antiderivative $\tfrac{2}{3}t^{3/2}$ at $4$ , not the integral.

G(4) = \int_4^4 \sqrt{t}\,dt = 0

❌ $\tfrac{1}{4}$ is the derivative of $\sqrt{t}$ at $4$ . FTC 2 gives

G'(x) = \sqrt{x}

itself.

❌ $G$ is zero at $4$ , but its slope is not. A function can pass through

0

while increasing;

G'(4) = \sqrt{4}

❌ Not quite. Use

G(a) = 0

and

G'(x) = f(x)

Show solution

G(4) = \int_4^4 \sqrt{t}\,dt = 0, \qquad G'(x) = \sqrt{x} \;\Longrightarrow\; G'(4) = 2.

Check with FTC 1: $G(x) = \tfrac{2}{3}x^{3/2} - \tfrac{16}{3}$ , so $G(4) = 0$ and $G'(x) = x^{1/2}$ .

Problem 3 · Both Theorems on One Function

Given: $G(x) = \displaystyle\int_1^x (3t^2 + 2t)\,dt$ . Find $G(3)$ and $G'(3)$ .

What is $G(3)$ ?

What is $G'(3)$ ?

✅ Correct!

G(x) = x^3 + x^2 - 2

, so

G(3) = 34

, and

G'(3) = 3\cdot 9 + 2\cdot 3 = 33

, straight from the integrand.

❌ The lower limit still counts.

t^3 + t^2

t = 1

2

; subtract it from

36

❌ $20$ is the derivative of the integrand, $6t + 2$ , at $3$ .

G'(3)

is the integrand itself at

3

❌ Check $G(3)$ . Evaluate

t^3 + t^2

from

t = 1

t = 3

❌ Check $G'(3)$ . FTC 2:

G'(x) = 3x^2 + 2x

Show solution

Value (FTC 1). An antiderivative of $3t^2 + 2t$ is $t^3 + t^2$ :

G(3) = \big(t^3 + t^2\big)\Big|_{t=1}^{t=3} = 36 - 2 = 34.

Slope (FTC 2). $G'(x) = 3x^2 + 2x$ , so $G'(3) = 27 + 6 = 33$ .

Check: $G(x) = x^3 + x^2 - 2$ has $G'(x) = 3x^2 + 2x$ ; the constant $-2$ washes away.

Problem 4 · Solving $y' = f(x)$

Find the solution of $y' = \dfrac{1}{1 + x^4}$ with $y(2) = 0$ .

✅ Correct! Its derivative is

\frac{1}{1+x^4}

by FTC 2, and at

x = 2

it is an integral over an interval of length zero.

❌ The derivative is right, but the starting value is not. At

x = 2

this is

\int_0^2 \frac{dt}{1+t^4} > 0

; the lower limit must be

2

❌ That is $f$ shifted, not a function whose derivative is $f$ . It vanishes at

2

, but its derivative is

-\frac{4x^3}{(1+x^4)^2}

❌ Reversed limits flip the sign.

\int_x^2 = -\int_2^x

, so its derivative is

-\frac{1}{1+x^4}

❌ Not quite. Use

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

with

a

equal to the starting point.

Show solution

Let $y(x) = \int_2^x \frac{dt}{1+t^4}$ . The integrand is continuous, so by FTC 2

y'(x) = \frac{1}{1 + x^4}, \qquad y(2) = \int_2^2 \frac{dt}{1+t^4} = 0.

Both conditions hold. No elementary formula is needed: Riemann sums compute $y(x)$ to any accuracy.

Solved: 0 / 4

Second Fundamental Theorem: Differentiate an Integral

IKey Formulas

IIVisualization 1 — The Area Grows at the Rate f(x)f(x)f(x)

IIIVisualization 2 — No +c+c+c: Moving the Lower Limit

IVQuiz Questions

IIVisualization 1 — The Area Grows at the Rate $f(x)$

IIIVisualization 2 — No $+c$ : Moving the Lower Limit