Fundamental theorem of calculus | Year 12

01 / LEARN

Understand the idea.

Accumulation and rate undo one another

An accumulation function has a variable endpoint. For $A (x) = \int_{a}^{x} f (t) d t$ , shifting $x$ adds or removes a narrow signed strip. If $f$ is continuous, the strip argument gives $A^{'} (x) = f (x)$ .

The integration letter $t$ is a dummy variable; after accumulation, the result depends on the endpoint $x$ . A constant lower bound sets the starting accumulation but contributes no derivative. You do not need to find an antiderivative first.

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

A moving endpoint needs the chain rule

If the upper endpoint is $g (x)$ , two rates combine: the integrand has value $f (g (x))$ at that endpoint, and the endpoint moves at rate $g^{'} (x)$ . Multiply them using the chain rule.

Keep this order clear: substitute the complete endpoint into the integrand, then multiply by its derivative. A decreasing endpoint may give a negative multiplier. The rule assumes the integrand is continuous over the relevant accumulation interval.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

If $A (x) = \int_{1}^{x} (t^{2} + 4) d t$ , find $A^{'} (x)$ .

Worked solution

1
Identify the endpoint and integrand.
$f (t) = t^{2} + 4, g (x) = x$
The polynomial is continuous and the lower endpoint 1 is fixed.
2
Differentiate the accumulation.
$A^{'} (x) = f (x) \cdot 1 = x^{2} + 4$
The fundamental theorem evaluates the integrand at the moving endpoint; its derivative is 1.
3
Verify by evaluating first.
$\begin{aligned} A (x) & = \frac{x^{3}}{3} + 4 x - \frac{13}{3} \\ \frac{d A}{d x} & = x^{2} + 4 \end{aligned}$
The fixed lower endpoint contributes only a constant, which disappears on differentiation.

Answer

$x^{2} + 4$

EXAMPLE 02

Worked example

Find $\frac{d}{d x} \int_{0}^{2 x} cos t d t$ .

Worked solution

1
Identify the moving endpoint.
$g (x) = 2 x, g^{'} (x) = 2$
The upper limit changes twice as fast as $x$ .
2
Evaluate the integrand at that endpoint.
$f (g (x)) = cos (2 x)$
The whole input $2 x$ replaces the integration variable $t$ .
3
Apply the chain-rule multiplier.
$\frac{d}{d x} \int_{0}^{2 x} cos t d t = 2 cos (2 x)$
The accumulation rate is endpoint height multiplied by endpoint speed.
4
Check using a direct antiderivative.
$\int_{0}^{2 x} cos t d t = sin (2 x) - sin 0 = sin (2 x)$
Differentiating this evaluated integral gives the same result.

Answer

$2 cos (2 x)$

03 / YOUR TURN

Try it yourself.

Check your understanding

Find $\frac{d}{d x} \int_{2}^{x} e^{t} d t$ .

Show answer & reasoning

$e^{x}$

The exponential is continuous, the lower endpoint 2 is fixed and the upper endpoint derivative is 1. Therefore $\frac{d}{d x} \int_{2}^{x} e^{t} d t = e^{x}$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

$A (x) = \int_{0}^{x} (3 t^{2} - 1) d t$ . Find $A^{'} (x)$ .
3 marks
$B (x) = \int_{1}^{x^{2}} t^{3} d t$ . Find $B^{'} (x)$ .
3 marks
$C (x) = \int_{0}^{3 x + 1} sin t d t$ . Find $C^{'} (x)$ .
3 marks
$D (x) = \int_{0}^{1 - x} e^{t} d t$ . Find $D^{'} (x)$ .
3 marks
$A (x) = \int_{0}^{x} (2 t + 1) d t$ . Evaluate $A$ first, then differentiate to check the theorem.
3 marks
$F^{'} (x) = f (x)$ and $F (2) - F (0) = 7$ . Find $\int_{0}^{2} f (x) d x$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$A (x) = \int_{0}^{x} (3 t^{2} - 1) d t$ . Find $A^{'} (x)$ .
3 marks
Apply the fundamental theorem directly.
$3 x^{2} - 1$
$B (x) = \int_{1}^{x^{2}} t^{3} d t$ . Find $B^{'} (x)$ .
3 marks
${(x^{2})}^{3} (2 x) = 2 x^{7}$ .
$2 x^{7}$
$C (x) = \int_{0}^{3 x + 1} sin t d t$ . Find $C^{'} (x)$ .
3 marks
Endpoint derivative 3 multiplies the sine value.
$3 sin (3 x + 1)$
$D (x) = \int_{0}^{1 - x} e^{t} d t$ . Find $D^{'} (x)$ .
3 marks
The endpoint derivative is $- 1$ .
$- e^{1 - x}$
$A (x) = \int_{0}^{x} (2 t + 1) d t$ . Evaluate $A$ first, then differentiate to check the theorem.
3 marks
${[t^{2} + t]}_{0}^{x} = x^{2} + x$ .
Differentiate and recover the integrand at $x$ .
$A = x^{2} + x$ and $A^{'} = 2 x + 1$
$F^{'} (x) = f (x)$ and $F (2) - F (0) = 7$ . Find $\int_{0}^{2} f (x) d x$ .
3 marks
The fundamental theorem gives the stated endpoint difference.
$7$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

In this lesson

Understand the idea.

Accumulation and rate undo one another

A moving endpoint needs the chain rule

See the method in action.

Worked example

Identify the endpoint and integrand.

Differentiate the accumulation.

Verify by evaluating first.

Worked example

Identify the moving endpoint.

Evaluate the integrand at that endpoint.

Apply the chain-rule multiplier.

Check using a direct antiderivative.

Try it yourself.

Check your understanding

Keep practising