AS & A Level · Practice questions · Definite integrals and area

Limits in, area out.

Six original Cambridge-style questions on definite integrals and areas, finishing with a short proof linking areas against both axes.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Sketch, set limits, integrate, subtract.

Sketch: where the region is, where the graphs meet, and whether it crosses the axis.
Limits: from the question or from solving the equations simultaneously.
Integrate: (upper \(-\) lower), or \(y\) for a region on the \(x\)-axis, or \(x\) in terms of \(y\) against the \(y\)-axis.
Evaluate: \(F(b) - F(a)\), with brackets around \(F(a)\).
Below the axis: integrate each part separately and add the sizes.

Recall

[2 marks]

Evaluate \(\displaystyle\int_{1}^{4} (2x + 3)\,dx\).

\(\left[x^{2} + 3x\right]_{1}^{4} = (16 + 12) - (1 + 3)\) ✓
\(= 28 - 4 = 24\) ✓

Calculation

[4 marks]

Find the exact value of \(\displaystyle\int_{1}^{4} \dfrac{x - 1}{\sqrt{x}}\,dx\).

\(\dfrac{x - 1}{\sqrt{x}} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\). ✓
Integrate: \(\left[\tfrac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}}\right]_{1}^{4}\). ✓
\(= \left(\tfrac{16}{3} - 4\right) - \left(\tfrac{2}{3} - 2\right) = \tfrac{4}{3} + \tfrac{4}{3}\). ✓
\(= \dfrac{8}{3}\) ✓

Calculation

[4 marks]

Evaluate (a) \(\displaystyle\int_{1}^{\infty} \dfrac{1}{(2x - 1)^{2}}\,dx\) (b) \(\displaystyle\int_{0}^{4} \dfrac{1}{\sqrt{x}}\,dx\).

(a) \(\displaystyle\int (2x - 1)^{-2}\,dx = \dfrac{(2x - 1)^{-1}}{-1 \times 2} = -\dfrac{1}{2(2x - 1)}\). ✓
As \(x \to \infty\) this tends to 0; at \(x = 1\) it is \(-\tfrac{1}{2}\). Value \(= 0 - \left(-\tfrac{1}{2}\right) = \tfrac{1}{2}\). ✓
(b) \(\left[2\sqrt{x}\right]_{0}^{4} = 4 - 0 = 4\). ✓

Analysis

[4 marks]

Find the area of the region enclosed by the curve \(y = 4x - x^{2}\) and the \(x\)-axis.

\(4x - x^{2} = x(4 - x) = 0\) at \(x = 0\) and \(x = 4\); the curve is above the axis between them. ✓
\(\displaystyle\int_{0}^{4} (4x - x^{2})\,dx = \left[2x^{2} - \tfrac{x^{3}}{3}\right]_{0}^{4}\). ✓
\(= 32 - \tfrac{64}{3} = \dfrac{32}{3}\) ✓

Analysis

[5 marks]

The curve \(y = x^{2} - 2x + 3\) and the line \(y = x + 3\) meet at two points.

(a) Find the \(x\)-coordinates of these points.

(b) Find the area of the region enclosed by the curve and the line.

(a) \(x^{2} - 2x + 3 = x + 3\) gives \(x^{2} - 3x = 0\), so \(x = 0\) or \(x = 3\). ✓
(b) Line minus curve: \((x + 3) - (x^{2} - 2x + 3) = 3x - x^{2}\). ✓
\(\displaystyle\int_{0}^{3} (3x - x^{2})\,dx = \left[\tfrac{3x^{2}}{2} - \tfrac{x^{3}}{3}\right]_{0}^{3} = \tfrac{27}{2} - 9\). ✓
\(= \dfrac{9}{2}\) ✓

Proof

[5 marks]

The rectangle \(OABC\) has vertices \(O(0, 0)\), \(A(2, 0)\), \(B(2, 4)\) and \(C(0, 4)\). The curve \(y = x^{2}\) passes through \(O\) and \(B\) and divides the rectangle into two regions.

(a) Find the area of the region between the curve and the \(x\)-axis.

(b) By integrating with respect to \(y\), find the area of the region between the curve and the \(y\)-axis.

(a) \(\displaystyle\int_{0}^{2} x^{2}\,dx = \left[\tfrac{x^{3}}{3}\right]_{0}^{2} = \dfrac{8}{3}\). ✓
(b) \(x = \sqrt{y}\): \(\displaystyle\int_{0}^{4} y^{\frac{1}{2}}\,dy = \left[\tfrac{2}{3}y^{\frac{3}{2}}\right]_{0}^{4} = \tfrac{2}{3} \times 8 = \dfrac{16}{3}\). ✓
(c) \(\tfrac{8}{3} + \tfrac{16}{3} = 8 = 2 \times 4\), the area of the rectangle, as it must be. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.