Evaluate _1^2 4x\,dx. [3 marks]

SQA SQA Higher 2022-style practice: Evaluate _0^π (2sin x + 3)\,dx, giving your answer in exact form.

Integrate term by term: (2sin x + 3)\,dx = -2cos x + 3x (2 marks). Apply the limits: [-2cos x + 3x]_0^π = (-2cosπ + 3π) - (-2cos 0 + 0) (1 mark). Use cosπ = -1 and cos 0 = 1: = (-2(-1) + 3π) - (-2(1)) = (2 + 3π) - (-2) = 4 + 3π (2 marks). Markers reward the integral of sin with the correct sign, substituting the exact trig values, and the simplified exact answer.

ScotlandMathsSyllabus dot point

How do you reverse differentiation to find a function from its gradient, and how does the definite integral measure area under a curve?

Integration as the reverse of differentiation, the indefinite integral of polynomial and trigonometric functions with the constant of integration, the definite integral, and the use of integration to find the area under a curve and the area between two curves.

A focused answer to the SQA Higher Mathematics integration content, covering integration as the reverse of differentiation, the indefinite integral of polynomial and trigonometric functions, the constant of integration, the definite integral, and finding the area under a curve and between two curves.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The indefinite integral
The definite integral and area
Integrating trigonometric functions
Examples in context
Try this

What this dot point is asking

The SQA wants you to integrate polynomial and trigonometric functions as the reverse of differentiation, include the constant of integration, evaluate definite integrals, and use integration to find the area under a curve and the area between two curves.

The indefinite integral

Integration is the reverse of differentiation: it answers "what function has this derivative". Because differentiating any constant gives zero, the reverse process must add an unknown constant $c$ to an indefinite integral.

The power rule for integration is the mirror of the one for differentiation: add one to the power and divide by the new power, then add $c$ . You can recover the value of $c$ when you are told a point the curve passes through.

Integrate

\displaystyle\int \left(4x^3 + \dfrac{1}{x^2}\right) dx

Find the indefinite integral, handling the reciprocal term correctly.

Step 1: Rewrite every term as a power of $x$

The power rule for integration requires the form $x^n$ . The fraction $\dfrac{1}{x^2}$ is $x^{-2}$ , so the integral becomes:

\int \left(4x^3 + x^{-2}\right) dx

Rewriting as a power is essential before integrating; you cannot apply "add one, divide by new power" to $\dfrac{1}{x^2}$ directly.

Step 2: Apply the power rule term by term

For each term, increase the power by one and divide by the new power:

\dfrac{4x^{3+1}}{4} + \dfrac{x^{-2+1}}{-2+1} + c = x^4 + \dfrac{x^{-1}}{-1} + c = x^4 - x^{-1} + c

Step 3: Write the answer in tidy form

Convert the negative power back to a fraction:

x^4 - \dfrac{1}{x} + c

Final answer: $\displaystyle\int \left(4x^3 + \dfrac{1}{x^2}\right) dx = x^4 - \dfrac{1}{x} + c$ .

The definite integral and area

A definite integral has limits, evaluates to a number, and needs no constant of integration because $c$ cancels in the subtraction. Substitute the upper limit into the integrated function, then the lower limit, and subtract.

Find the area under

y = x^2

between

x = 0

and

x = 3

The curve $y = x^2$ lies above the x-axis for $x \ge 0$ , so the definite integral gives the area directly.

Step 1: Set up the definite integral

The area between the curve and the x-axis from $x = 0$ to $x = 3$ is:

\int_0^3 x^2\,dx

Unlike an indefinite integral, a definite integral has fixed limits and no constant of integration (it cancels in the subtraction).

Step 2: Integrate using the power rule

Apply "add one to the power and divide" to get the integrated function in square brackets:

\int_0^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^3

Step 3: Substitute the limits and subtract

Evaluate at the upper limit, then subtract the value at the lower limit:

\dfrac{3^3}{3} - \dfrac{0^3}{3} = \dfrac{27}{3} - 0 = 9

Final answer: The area under $y = x^2$ between $x = 0$ and $x = 3$ is $9$ square units.

For the area between two curves, integrate the difference (upper curve minus lower curve) between the points where they meet.

Evaluate the definite integral

\displaystyle\int_1^4 \left(2x - \dfrac{3}{\sqrt{x}}\right) dx

Evaluate the definite integral, first rewriting the surd term as a power so the standard rule applies.

Step 1: Rewrite the surd term as a fractional power

The rule $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ requires the form $x^n$ . Write $\dfrac{3}{\sqrt{x}} = 3x^{-1/2}$ , so the integrand becomes $2x - 3x^{-1/2}$ .

Step 2: Integrate each term

Apply the power rule to each term. For $2x = 2x^1$ : divide to get $x^2$ . For $-3x^{-1/2}$ : increase the power to $\dfrac{1}{2}$ , then divide by $\dfrac{1}{2}$ (which means multiplying by $2$ ):

\int \left(2x - 3x^{-1/2}\right) dx = x^2 - 3 \cdot \dfrac{x^{1/2}}{\frac{1}{2}} = x^2 - 6x^{1/2} = x^2 - 6\sqrt{x}

Step 3: Apply the limits

Evaluate at $x = 4$ : $16 - 6\sqrt{4} = 16 - 12 = 4$ . Evaluate at $x = 1$ : $1 - 6\sqrt{1} = 1 - 6 = -5$ . Subtract:

\left[x^2 - 6\sqrt{x}\right]_1^4 = 4 - (-5) = 9

Final answer: The value of the integral is $9$ .

Integrating trigonometric functions

The trigonometric integrals reverse the derivatives, so a sign appears for $\sin$ . A common slip is to forget that $\displaystyle\int \sin x\,dx = -\cos x + c$ carries a minus sign while $\displaystyle\int \cos x\,dx = \sin x + c$ does not.

Find

\displaystyle\int (\cos x - 2\sin x)\,dx

Integrate the trigonometric expression, taking care with the sign that appears when integrating sine.

Step 1: Recall the trigonometric integration rules

The key facts are $\displaystyle\int \cos x\,dx = \sin x + c$ and $\displaystyle\int \sin x\,dx = -\cos x + c$ . The minus sign in the second rule is easily forgotten: it arises because differentiating $-\cos x$ gives $\sin x$ .

Step 2: Integrate each term

For the $\cos x$ term: $\displaystyle\int \cos x\,dx = \sin x$ . For the $-2\sin x$ term: $\displaystyle\int -2\sin x\,dx = -2(-\cos x) = 2\cos x$ .

Step 3: Combine and add the constant

\int (\cos x - 2\sin x)\,dx = \sin x + 2\cos x + c

Final answer: $\sin x + 2\cos x + c$ . You can verify by differentiating: $\dfrac{d}{dx}(\sin x + 2\cos x) = \cos x - 2\sin x$ , which recovers the original integrand.

Examples in context

Integration recovers a total from a rate. If water flows into a reservoir at a rate $r(t) = 6t$ litres per hour, the volume added between hours $1$ and $3$ is $\displaystyle\int_1^3 6t\,dt = \big[3t^2\big]_1^3 = 27 - 3 = 24$ litres. The same machinery that measures area under the rate-time graph measures the accumulated volume, which is why definite integration models any total-from-a-rate problem.

Try this

Q1. Find $\displaystyle\int (6x^2 - 4)\,dx$ . [2 marks]

Cue. $2x^3 - 4x + c$ .

Q2. Evaluate $\displaystyle\int_1^2 4x\,dx$ . [3 marks]

Cue. $[2x^2]_1^2 = 8 - 2 = 6$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20194 marksA curve passes through the point

(1, 4)

and has

\dfrac{dy}{dx} = 6x^2 - 2x

. Express

y

in terms of

x

Show worked answer →

Integrate the gradient function: $y = \displaystyle\int (6x^2 - 2x)\,dx = 2x^3 - x^2 + c$ (2 marks, including the constant).

Use the point $(1, 4)$ to find $c$ : $4 = 2(1)^3 - (1)^2 + c = 2 - 1 + c = 1 + c$ , so $c = 3$ (1 mark).

Therefore $y = 2x^3 - x^2 + 3$ (1 mark). Markers reward the correct integral with $+ c$ , substituting the point to find $c$ , and the final expression.

SQA Higher 20225 marksEvaluate

\displaystyle\int_0^{\pi} (2\sin x + 3)\,dx

, giving your answer in exact form.

Show worked answer →

Integrate term by term: $\displaystyle\int (2\sin x + 3)\,dx = -2\cos x + 3x$ (2 marks).

Apply the limits: $\big[-2\cos x + 3x\big]_0^{\pi} = \big(-2\cos\pi + 3\pi\big) - \big(-2\cos 0 + 0\big)$ (1 mark).

Use $\cos\pi = -1$ and $\cos 0 = 1$ : $= \big(-2(-1) + 3\pi\big) - \big(-2(1)\big) = (2 + 3\pi) - (-2) = 4 + 3\pi$ (2 marks). Markers reward the integral of $\sin$ with the correct sign, substituting the exact trig values, and the simplified exact answer.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)