The Fundamental Theorem of Calculus

Linking differentiation and integration through the evaluation of definite integrals.

Core Theorem

Part1: The Derivative of an Integral:

\frac{d}{dx} \int_{a}^{g(x)} f(t) dt = f(g(x)) \cdot g'(x)

Part2 : The Evaluation of Definite Integrals:

\int_{a}^{b} f(x) dx = F(b) - F(a)

, where

F'(x) = f(x)

.
Total Change / Accumulation:In AP Calculus FRQs, FTC is often expressed as:

f(b) = f(a) + \int_{a}^{b} f'(x) dx

. Current State = Initial State + Accumulated Change.

Step-by-Step SOP

1
Differentiation Procedure (FTC 1)
1. Identify the upper limit $g(x)$ . 2. Substitute $g(x)$ into the integrand $f(t)$ . 3. Multiply the result by $g'(x)$ .
2
Evaluation Procedure (FTC 2)
1. Find the antiderivative $F(x)$ . 2. Substitute the upper limit $b$ and lower limit $a$ . 3. Compute the difference: $F(b) - F(a)$ .
3
Real-World Application (Accumulation)
To find a final value, use: $Final = Initial + \int_{start}^{end} Rate \, dt$ .

Practice Exercises

Example 01Medium

Find

h'(x)

h(x) = \int_{0}^{x^2} \sin(t) dt

NEED A HINT?

Use FTC Part 1 with the Chain Rule.

SHOW DETAILED EXPLANATION

Plug the upper limit

x^2

into the function:

\sin(x^2)

Multiply by the derivative of the upper limit:

\frac{d}{dx}(x^2) = 2x

h'(x) = 2x \sin(x^2)

Example 02Easy

Evaluate

\int_{1}^{3} x^2 dx

NEED A HINT?

Use FTC Part 2.

SHOW DETAILED EXPLANATION

F(x) = \frac{1}{3}x^3

Calculate

F(3) - F(1) = \frac{1}{3}(3)^3 - \frac{1}{3}(1)^3 = 9 - \frac{1}{3} = \frac{26}{3}

Common Pitfalls

⚠
Forgetting the Chain RuleWhen the upper limit is a function like $x^2$ instead of just $x$ , you must multiply by its derivative ( $2x$ ).
⚠
Lower Limit Constant ConfusionIn FTC Part 1, if the lower limit 'a' is a constant, it does not affect the derivative. Don't try to subtract it during the differentiation process.
⚠
Sign Errors in EvaluationAlways use parentheses when subtracting the lower limit value: $F(b) - (F(a))$ , especially if $F(a)$ contains multiple terms or negative signs.

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