Fundamental theorem of calculus | Year 12

01 / LEARN

The idea, step by step

Accumulation and rate undo one another

If $A (x) = \int_{a}^{x} f (t) d t$ and $f$ is continuous, increasing $x$ adds a thin strip whose height is approximately $f (x)$ . In the limit, $A^{'} (x) = f (x)$ . The letter $t$ inside the integral is a dummy variable; the upper endpoint $x$ determines the resulting function.

This result links the area/accumulation meaning of an integral with the antiderivative rules used to evaluate it. Do not differentiate the dummy variable as though it remained a free variable after integration.

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

A moving endpoint needs the chain rule

For $A (x) = \int_{a}^{g (x)} f (t) d t$ , the endpoint changes at rate $g^{'} (x)$ . The accumulation therefore changes at $f (g (x)) g^{'} (x)$ . A negative endpoint derivative can reverse the sign.

You can check this on a simple polynomial by first evaluating the integral and then differentiating. Both routes must agree. Use the theorem to avoid unnecessary integration when the integrand has no simple elementary antiderivative.

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

A common trap

Evaluating at the endpoint but forgetting the endpoint derivative.

Instead: Apply the chain rule after substituting the endpoint into the integrand.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

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

The upper endpoint is $x$ and the integrand is continuous.
Replace the dummy variable $t$ by $x$ ; no antiderivative is needed.

x^{2} + 4

Example 2

Worked example

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

The endpoint is $2 x$ , so use its derivative 2.
Multiply the integrand evaluated at $2 x$ by 2.

2 cos (2 x)

Check your understanding

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

Show answer

$e^{x}$

The lower limit is constant and the upper endpoint has derivative 1.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF 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 ↗

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