Find (4x^3 - 6x + 1)\,dx. [2 marks]

Evaluate _0^π sin x\,dx. [2 marks]

OCR OCR 2022-style practice: Use the trapezium rule with four strips to estimate _0^2 sqrt(4 + x^3)\,dx, giving your answer to three decimal places.

Four strips on [0, 2] give h = 2 - 04 = 0.5, with ordinates at x = 0, 0.5, 1, 1.5, 2 (M1). Evaluate y = sqrt(4 + x^3): y_0 = 2, y_1 = sqrt(4.125) ≈ 2.0310, y_2 = sqrt(5) ≈ 2.2361, y_3 = sqrt(7.375) ≈ 2.7157, y_4 = sqrt(12) ≈ 3.4641 (A1). Apply the rule (M1): int ≈ 0.52[y_0 + y_4 + 2(y_1 + y_2 + y_3)] = 0.25[5.4641 + 2(6.9828)] (A1). = 0.25[5.4641 + 13.9656] = 0.25 × 19.4297 ≈ 4.857 (A1). Markers reward the strip width, the ordinates, doubling only the interior ordinates, and the final estimate.

EnglandMathsSyllabus dot point

How do you reverse differentiation to find areas under and between curves, and integrate the standard functions?

Indefinite and definite integrals as the reverse of differentiation, the integrals of standard functions, the area under a curve and between two curves, and the trapezium rule for numerical integration.

A focused answer to the OCR A-Level Mathematics A integration content, covering indefinite and definite integrals as the reverse of differentiation, the integrals of standard functions, the area under a curve and between two curves, and the trapezium rule for estimating an integral numerically.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Quick answer

Integration reverses differentiation: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ (with the special case $\int x^{-1}\,dx = \ln|x| + c$ ), and always add the constant for an indefinite integral. A definite integral evaluates the antiderivative at the limits and subtracts. The area under a curve is the definite integral (take the magnitude where the curve dips below the axis), and the area between curves integrates upper minus lower between the intersection points. The trapezium rule $\frac{h}{2}[y_0 + y_n + 2(y_1 + \dots + y_{n-1})]$ estimates an integral, overestimating for a concave-up curve and underestimating for a concave-down one.

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants you to integrate as the reverse of differentiation (with a constant of integration for an indefinite integral), evaluate definite integrals, integrate the standard functions, find the area under a curve and the area between two curves, handle regions below the $x$ -axis, and estimate an integral numerically with the trapezium rule.

The answer

Integration as anti-differentiation

Integration reverses differentiation. For an indefinite integral always add the constant of integration $c$ .

The case $n = -1$ is special: $\displaystyle\int x^{-1}\,dx = \ln|x| + c$ , not a power.

Definite integrals

A definite integral evaluates the antiderivative at the limits and subtracts. No constant is needed because it cancels.

Area under and between curves

The area under $y = f(x)$ from $x = a$ to $x = b$ (where the curve is above the axis) is $\displaystyle\int_a^b f(x)\,dx$ . For a region below the axis the integral is negative, so take its magnitude. For the area between two curves, integrate the difference "upper minus lower".

The trapezium rule

When a function cannot be integrated exactly, estimate the area by dividing into strips of equal width and treating each top as a straight line.

Examples in context

Area between a line and a curve

Find the area enclosed between

y = x + 2

and

y = x^2

step 1 Find the intersections

$x^2 = x + 2$ gives $x^2 - x - 2 = (x - 2)(x + 1) = 0$ , so $x = -1$ and $x = 2$ .

step 2 Integrate upper minus lower

Between the roots the line is above the parabola, so area $= \displaystyle\int_{-1}^{2}\big[(x + 2) - x^2\big]\,dx = \left[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\right]_{-1}^{2}$ .

step 3 Evaluate

At $x = 2$ : $2 + 4 - \tfrac{8}{3} = 6 - \tfrac{8}{3} = \tfrac{10}{3}$ . At $x = -1$ : $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$ . Area $= \tfrac{10}{3} - (-\tfrac{7}{6}) = \tfrac{20}{6} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}$ .

Is the trapezium estimate too big or too small?

For a curve that bends upward (concave up) over the interval, each straight chord lies above the curve, so the trapezium rule overestimates. For a concave-down curve it underestimates. Increasing the number of strips improves the estimate.

Try this

Q1. Find $\displaystyle\int (4x^3 - 6x + 1)\,dx$ . [2 marks]

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

Q2. Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$ . [2 marks]

Cue. $[-\cos x]_0^{\pi} = -\cos\pi - (-\cos 0) = 1 + 1 = 2$ .

Exam-style practice questions

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

OCR 20196 marksThe region

R

is bounded by the curve

y = x^2 - 2x - 3

and the

x

-axis. Find the exact area of

R

Show worked answer →

Find where the curve meets the axis: $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$ , so $x = -1$ and $x = 3$ (M1, A1).

Between these roots the curve is below the axis, so the integral is negative and the area is its magnitude. Integrate (M1): $\int_{-1}^{3}(x^2 - 2x - 3)\,dx = \left[\dfrac{x^3}{3} - x^2 - 3x\right]_{-1}^{3}$ (A1).

At $x = 3$ : $9 - 9 - 9 = -9$ . At $x = -1$ : $-\tfrac{1}{3} - 1 + 3 = \tfrac{5}{3}$ . The integral is $-9 - \tfrac{5}{3} = -\tfrac{32}{3}$ (M1).

Area $= \left|-\tfrac{32}{3}\right| = \dfrac{32}{3}$ square units (A1).

Markers reward the roots, integrating, evaluating at the limits, and taking the magnitude for an area below the axis.

OCR 20225 marksUse the trapezium rule with four strips to estimate

\int_0^2 \sqrt{4 + x^3}\,dx

, giving your answer to three decimal places.

Show worked answer →

Four strips on $[0, 2]$ give $h = \dfrac{2 - 0}{4} = 0.5$ , with ordinates at $x = 0, 0.5, 1, 1.5, 2$ (M1).

Evaluate $y = \sqrt{4 + x^3}$ : $y_0 = 2$ , $y_1 = \sqrt{4.125} \approx 2.0310$ , $y_2 = \sqrt{5} \approx 2.2361$ , $y_3 = \sqrt{7.375} \approx 2.7157$ , $y_4 = \sqrt{12} \approx 3.4641$ (A1).

Apply the rule (M1): $\int \approx \dfrac{0.5}{2}\big[y_0 + y_4 + 2(y_1 + y_2 + y_3)\big] = 0.25\big[5.4641 + 2(6.9828)\big]$ (A1).

$= 0.25[5.4641 + 13.9656] = 0.25 \times 19.4297 \approx 4.857$ (A1).

Markers reward the strip width, the ordinates, doubling only the interior ordinates, and the final estimate.

Related dot points

Sources & how we know this

OCR A Level Mathematics A (H240) specification — OCR (2017)