Direct integration | Year 12

01 / LEARN

The idea, step by step

Integrate a known rate

When $d y / d x = f (x)$ , take an antiderivative of the right side and add a constant. This is direct integration because the right side contains no unknown $y$ . Use the power, exponential, trigonometric or logarithmic integration rule that matches the integrand.

For example $d y / d x = 3 x^{2} + e^{x}$ gives $y = x^{3} + e^{x} + C$ . The initial value selects $C$ . Differentiate the completed answer to check the equation rather than trusting the integration alone.

Accumulation and motion

The change in a quantity from $a$ to $b$ is the definite integral of its rate. If the initial amount is $y (a)$ , then $y (b) = y (a) + \int_{a}^{b} f (x) d x$ . A negative rate contributes negative change.

For motion, integrate acceleration to obtain velocity, then integrate velocity to obtain position. Each integration introduces its own constant, determined by the matching velocity or position condition. Displacement is signed; total distance requires integrating the magnitude of velocity and splitting at sign changes.

A common trap

Use the same arbitrary constant after both integrations.

Instead: Acceleration-to-velocity and velocity-to-position each introduce an independent constant.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Solve $y^{'} = 3 x^{2} + 2 e^{x}$ with $y (0) = 5$ .

An antiderivative is $x^{3} + 2 e^{x} + C$ .
$5 = 0 + 2 + C$ , so $C = 3$ .
The derivative returns the original right side.

y = x^{3} + 2 e^{x} + 3

Example 2

Worked example

A particle has $a (t) = 6 t$ , $v (0) = 2$ , $s (0) = 1$ . Find $s (t)$ .

Integrate acceleration: $v = 3 t^{2} + C_{1}$ . The condition gives $C_{1} = 2$ .
Integrate again: $s = t^{3} + 2 t + C_{2}$ .
$s (0) = 1$ gives $C_{2} = 1$ .

s = t^{3} + 2 t + 1

Check your understanding

Solve $y^{'} = 4 cos (2 x)$ , $y (0) = 1$ .

Show answer

$y = 2 sin (2 x) + 1$

Integrate with the factor $1 / 2$ , then apply the initial condition.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $y^{'} = 5 x^{4} - 2$ with $y (1) = 3$ .
2 marks
Solve $y^{'} = 6 e^{2 x}$ with $y (0) = 7$ .
2 marks
Solve $y^{'} = 2 / x$ with $y (1) = 4$ , $x > 0$ .
2 marks
A tank initially contains 20 L. The net inflow is $3 t + 2$ L/min for $0 \leq t \leq 4$ . Find its amount at 4 min.
2 marks
A particle has $a = 4$ , $v (0) = - 3$ , $s (0) = 2$ . Find $s (t)$ .
2 marks
A particle has $v (t) = t - 2$ m/s on $0 \leq t \leq 4$ . Find displacement and total distance.
2 marks

Read the worked solutions

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

Solve $y^{'} = 5 x^{4} - 2$ with $y (1) = 3$ .
2 marks
Integrate to $x^{5} - 2 x + C$ .
$3 = 1 - 2 + C$ , so $C = 4$ .
$y = x^{5} - 2 x + 4$
Solve $y^{'} = 6 e^{2 x}$ with $y (0) = 7$ .
2 marks
The antiderivative is $3 e^{2 x} + C$ .
$7 = 3 + C$ gives $C = 4$ .
$y = 3 e^{2 x} + 4$
Solve $y^{'} = 2 / x$ with $y (1) = 4$ , $x > 0$ .
2 marks
Integrate on the positive domain.
$ln 1 = 0$ , so the constant is 4.
$y = 2 ln x + 4$
A tank initially contains 20 L. The net inflow is $3 t + 2$ L/min for $0 \leq t \leq 4$ . Find its amount at 4 min.
2 marks
The increase is $\int_{0}^{4} (3 t + 2) d t = {[1.5 t^{2} + 2 t]}_{0}^{4} = 32$ L.
Add the initial 20 L.
$52$ L
A particle has $a = 4$ , $v (0) = - 3$ , $s (0) = 2$ . Find $s (t)$ .
2 marks
Integrate: $v = 4 t - 3$ .
Integrate again and use $s (0) = 2$ .
$s = 2 t^{2} - 3 t + 2$
A particle has $v (t) = t - 2$ m/s on $0 \leq t \leq 4$ . Find displacement and total distance.
2 marks
$\int_{0}^{4} (t - 2) d t = 0$ .
The velocity changes sign at $t = 2$ . The two triangular speed areas are each 2 m.
Displacement $0$ m; total distance $4$ m.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

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