Direct integration | Year 12

01 / LEARN

Understand the idea.

Integrate a known rate

When $y^{'} = f (x)$ contains no unknown $y$ on the right, direct integration gives $y = \int f (x) d x + C$ . Choose the integration rule term by term, retaining coefficients and any chain-rule scale factors.

Use the initial condition only after obtaining the antiderivative family. Determine its constant by substitution, then differentiate the completed function. This checks the integration, while a second substitution checks that the correct member of the family was selected.

Accumulation and motion

A rate accumulates to a change: $y (b) - y (a) = \int_{a}^{b} f (x) d x$ . Add the initial quantity if the final quantity is required. A negative rate reduces the accumulation, so the result need not be positive.

In motion, integrate acceleration to find velocity, then velocity to find position. Each integration introduces its own constant, matched to its own initial condition. Displacement is signed; total distance integrates $∣ v ∣$ and may require splitting where velocity changes sign.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Integrate the two terms.
$y = \int (3 x^{2} + 2 e^{x}) d x = x^{3} + 2 e^{x} + C$
The right side depends only on x, so its antiderivative gives the solution family.
2
Determine the constant.
$5 = 0^{3} + 2 e^{0} + C = 2 + C, C = 3$
The initial output includes the exponential contribution as well as C.
3
State and check the solution.
$\begin{array}{c} y = x^{3} + 2 e^{x} + 3 \\ y^{'} = 3 x^{2} + 2 e^{x}, y (0) = 5 \end{array}$
Both the differential equation and the initial condition are satisfied.

Answer

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

EXAMPLE 02

Worked example

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

Worked solution

1
Integrate acceleration to get velocity.
$v (t) = \int 6 t d t = 3 t^{2} + C_{1}$
Acceleration is the time derivative of velocity.
2
Use the velocity condition.
$v (0) = 2 \Rightarrow C_{1} = 2, v (t) = 3 t^{2} + 2$
The initial velocity fixes the first integration constant.
3
Integrate velocity and use the position condition.
$\begin{array}{c} s (t) = \int (3 t^{2} + 2) d t = t^{3} + 2 t + C_{2} \\ s (0) = 1 \Rightarrow C_{2} = 1 \end{array}$
The position condition fixes a separate constant from the velocity condition.
4
Verify the position model.
$s = t^{3} + 2 t + 1, s^{'} = 3 t^{2} + 2, s^{′′} = 6 t$
Differentiating twice recovers acceleration, and the initial position and velocity have the required values.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

Integrating $4 cos (2 x)$ gives $2 sin (2 x) + C$ because differentiation supplies a factor 2. Since $sin 0 = 0$ , the condition $y (0) = 1$ gives $C = 1$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $y^{'} = \frac{2}{x}$ 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 $y^{'} = \frac{2}{x}$ 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 ↗

In this lesson

Understand the idea.

Integrate a known rate

Accumulation and motion

See the method in action.

Worked example

Integrate the two terms.

Determine the constant.

State and check the solution.

Worked example

Integrate acceleration to get velocity.

Use the velocity condition.

Integrate velocity and use the position condition.

Verify the position model.

Try it yourself.

Check your understanding

Keep practising