Further Integration (Cambridge International A-Level Mathematics): Exam Questions

The area bounded by the curve with equation $y = 9 - x^{2}$, the $x$-axis and the vertical lines with equations $x = 1$ and $x = 2$ is to be found.

Write down an integral that would find this area.

Evaluate your integral from part (a) and hence find the area described above.

The diagram below shows the graph of $y = 7 x - x^{2} - 6$.

Find the shaded area, giving your answer as a fraction in its simplest terms.

The diagram below shows the graph of $y = x^{2} - 8 x + 12$.

Find the shaded area marked $R$, giving your answer as a fraction in its simplest form.

Simplify x - \left(x^{2} - 8x + 18\right).

The diagram below shows the graphs of $y = x^{2} - 8 x + 18$ and $y = x$.

Find the shaded area marked $R$, giving your answer as a fraction in its simplest terms.

The diagram below shows the region $R$, bounded by the straight lines with equations $y = 2 x + 1$, $x = 2$, $x = 4$ and the $x$-axis.

(i) For $y = 2 x + 1$, show that $y^{2} = 4 x^{2} + 4 x + 1$.

(ii) Find the volume of the solid formed when the region $R$ is rotated 360° around the $x$-axis.

The diagram shows the curve with equation $y = x^{2} + 2$. The shaded region $R$ is bounded by the curve, the $y$-axis and the lines $y = 6$ and $y = 11$.

Find the volume of the solid formed when the shaded region $R$ is rotated through 360° about the $y$-axis.

The diagram below shows part of the graph of $y = x^{2} - 4 x + 3$.

Find the area of the shaded region labelled $R$.

The diagram below shows part of the graph of y = 4 - \left(x - 2\right)^{2}.

Write down the values of $x$ where $y = 0$.

Show that

4 - \left(x - 2\right)^{2} = 4x - x^{2}

Evaluate

\int_{0}^{4} \left(4x - x^{2}\right) \,\text{d}x

Write down the area of the region labelled $R$.

Find the $x$-coordinates of the points of intersection of the line with equation $y = 2$ and the curve with equation $y = x^{2} - 4 x + 5$.

Evaluate

\int_{1}^{3} \left(x^{2} - 4x + 5\right) \,\text{d}x

The diagram below shows the graphs of $y = 2$ and $y = x^{2} - 4 x + 5$.

Find the exact area of the shaded region $R$.

The diagram below shows the graphs of the line $y = 6 - x$ and the curve $y = x^{2}$.

Find the $x$-coordinates of the points labelled $P$, $Q$ and $R$.

Find the area of the shaded region.

The diagram below shows a sketch of the curves with equations

$y = x^{2} - 3 x + 4$ and $y = 4 - x^{2} + 2 x$

Find the $x$-coordinates of the points of intersection of the two graphs.

Show that the area of the shaded region labelled $R$ is given by

\int_{0}^{\frac{5}{2}} \left(5x - 2x^{2}\right) \,\text{d}x

Use calculus to find the area of the shaded region labelled $R$.

The diagram below shows the graph of the curve with equation $y = 4 - x^{2}$.

(i) Find the $x$-coordinates of the points where the graph of $y = 4 - x^{2}$ crosses the $x$-axis.

(ii) The shaded region $R$ is rotated through 360° about the $x$-axis. Find the volume of the solid formed.

The diagram below shows the graphs of two horizontal lines and the function $y = \text{f} ( x )$.

$\text{f} ( x ) = a x^{2} + b$, where $a$ and $b$ are constants.

Point $P$ has coordinates \left(1, 5\right).

Point $Q$ has coordinates \left(2, 8\right).

(i) Find the values of $a$ and $b$.

(ii) Write down the equations of the two horizontal lines.

The shaded region $R$ is rotated through 360° about the $y$-axis. Find the volume of the solid formed.

The diagram below shows part of the graph of $y = 2 x + 3 x^{2} - 2 x^{3}$.

Find the area of the shaded region labelled $R$.

The diagram below shows part of the graph of y = x\left(x - 1\right)\left(x + 2\right).

Find the total area of the two shaded regions.

The line with equation $5 y = 14 - 3 x$ cuts the curve with equation $5 y = 20 - 2 x - x^{2}$ at the points $P$ and $Q$, as shown.

Find the $x$- and $y$-coordinates of the points $P$ and $Q$.

Find the exact area of the region labelled $R$, giving your answer in the form $\frac{a}{b}$, where $a$ and $b$ are integers to be found.

The diagram below shows the graphs of $y = x + 2$ and $y = 10 x - x^{2} - 16$.

Find the exact area of the shaded region.

The diagram below shows a sketch of the curves with equations

$y = x^{3} + 3$ and $y = - x^{3} + 2 x + 3$

Find the $x$-coordinates of the points of intersection of the two graphs.

Use calculus to find the total shaded area enclosed by the two graphs.

A mathematical model for a bowl is obtained by rotating through 360° about the $y$-axis the part of the curve $y = \frac{1}{a} x^{2}$ which is between $y = 8$ and $y = 9$, and then adding a flat bottom. $a$ is a constant such that $a > 0$.

Find the capacity of the bowl in terms of $a$.

In the case where the boundaries $y = 3$ and $y = 6$ are used in place of $y = 8$ and $y = 9$, the capacity of the bowl is increased by $15 π$ cubic units.

Find the value of $a$.

What assumption has been made in finding the capacity of the bowl?

The diagram below shows a right-angled triangle with vertices at the origin, the point \left(h, 0\right) and the point \left(h, r\right), where $r > 0$ and $h > 0$.

Find an equation of the line on which the hypotenuse of the right-angled triangle lies, giving your answer in the form $y = \text{f} ( x )$.

The triangle is rotated through 360° about the $x$-axis to form a cone.

Hence use calculus to show that the general formula for the volume $V$ of a cone is

$V = \frac{1}{3} π r^{2} h$

where $r$ is the base radius of the cone and $h$ is its perpendicular height.

The diagram below shows part of the curve $C$ defined by the equation $y = \frac{1}{a} x^{2}$, where $a$ is a positive constant. The shaded region $R$ is bounded by the curve, the $x$-axis, and the lines $x = 1$ and $x = 6$.

Given that the volume of the solid formed when the region $R$ is rotated through 360° about the $x$-axis is $\frac{311π}{20}$ cubic units, find the value of $a$.

The diagram below shows part of the graph of $y = x^{3} - 2 x^{2} - x + 2$.

Find the total area of the two shaded regions.

The diagram below shows the graphs of $y = 12 x - x^{2} - 27$ and $y = x^{2} - 14 x + 45$.

Find the area of the shaded region, $R$.

Starting with the equation of a semicircle of radius $r$, $y = \sqrt{r^{2}-x^{2}}$ (where $r > 0$), use calculus to show that the general formula for the volume $V$ of a sphere of radius $r$ is

$V = \frac{4}{3} π r^{3}$

The diagram below shows part of the graph of $y = 4 x - x^{3}$.

Find the area of the shaded region labelled $R$.

Without doing any additional calculation, explain why \int_{-2}^{2} \left(4x - x^{3}\right) \,\text{d}x must be equal to zero.

The diagram below shows the graphs of $4 y = 4 x + 17$ and $y = 3 x^{2} - 8 x - 16$.

Find the exact area of the shaded region.

Sketch the region bounded by the lines

$x = 2$, $x = 6$, $2 y = x + 4$ and $y = p$, where $0 < p < 2$.

The region described in part (a) is rotated through 360° about the $x$-axis.

Find the volume of the solid formed, giving your answer in terms of $p$.

The solid formed in part (b) will have a 'hole' in its centre.

(i) Find the volume of this 'hole', giving your answer in terms of $p$.

(ii) Hence show that there are no values of $p$ in the given interval that make the volume of the solid equal to the volume of the 'hole'.

The diagram below shows a sketch of the curves with equations $y = x^{2} + 5$ and $y = 23 - x^{2}$.

The region $R$ is bounded by the $y$-axis and the two curves. This region is rotated through 360° about the $y$-axis. Find the volume of the solid formed.

The regions marked $R$ and $S$ are now to be considered as a single region bounded by the two curves.

A student calculates the volume swept out when this combined region is rotated through a full 360° about the $y$-axis, without taking the symmetry of the two curves into account. Call this value $I$.

How does $I$ compare to the answer you found in part (a)?

Be sure to fully explain your answer.