Areas and volumes
Integration
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Find the area enclosed by the curve $y = x^2$, the $x$-axis and the lines $x = 0$ and $x = 3$.
3 marks
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Find the area enclosed by the curves $y = x$ and $y = x^2$ between their points of intersection.
3 marks
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Find the exact area enclosed by the curve $y = x^2 - 4$ and the $x$-axis.
4 marks
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The region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid formed.
4 marks
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Find the exact area of the finite region enclosed by $y = x^2$ and $y = x + 2$.
5 marks
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The region under $y = \dfrac{1}{x}$ from $x = 1$ to $x = 3$ is rotated through $360^\circ$ about the $x$-axis. Find the exact volume.
5 marks
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The region under $y = \sqrt{x}$ from $x = 1$ to $x = 4$ is rotated through $360^\circ$ about the $x$-axis. Find the exact area of the region and the exact volume of the solid.
6 marks
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The finite region enclosed by $y = x^2$ and $y = x + 2$ is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid formed.
7 marks
Mark scheme
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Answer. $9$
- $\displaystyle \int_0^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^3 = 9$.
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Answer. $\dfrac{1}{6}$
- The curves meet at $x = 0$ and $x = 1$.
- $\displaystyle \int_0^1 (x - x^2)\,dx = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$.
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Answer. $\dfrac{32}{3}$
- The curve meets the axis at $x = \pm 2$ and lies below it between those points.
- Area $= \displaystyle \int_{-2}^{2} (4 - x^2)\,dx = \dfrac{32}{3}$.
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Answer. $8\pi$
- Volume $= \pi \displaystyle \int_0^4 x\,dx = \pi\left[\dfrac{x^2}{2}\right]_0^4 = 8\pi$.
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Answer. $\dfrac{9}{2}$
- The curves meet when $x^2 - x - 2 = 0$, so $x = -1$ or $x = 2$.
- $\displaystyle \int_{-1}^{2} (x + 2 - x^2)\,dx = \dfrac{9}{2}$.
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Answer. $\dfrac{2\pi}{3}$
- Volume $= \pi \displaystyle \int_1^3 x^{-2}\,dx = \pi\left[-\dfrac{1}{x}\right]_1^3 = \dfrac{2\pi}{3}$.
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Answer. The area is $\dfrac{14}{3}$ and the volume is $\dfrac{15\pi}{2}$.
- Area $= \displaystyle \int_1^4 x^{1/2}\,dx = \dfrac{14}{3}$.
- Volume $= \pi \displaystyle \int_1^4 x\,dx = \dfrac{15\pi}{2}$.
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Answer. $\dfrac{72\pi}{5}$
- The curves meet at $x = -1$ and $x = 2$.
- Volume $= \pi \displaystyle \int_{-1}^{2} \big((x + 2)^2 - x^4\big)\,dx$.
- The integral of $-x^4 + x^2 + 4x + 4$ between these limits is $\dfrac{72}{5}$.
- The volume is $\dfrac{72\pi}{5}$.