Integration

A curve whose gradient is given by $\frac{dy}{dx} = 3x^2 - \frac{2}{x^3}$ passes through $(-1, 3)$.

Feb/March 2016

The diagram depicts the curve $y = f(x)$ for $x > 0$. It has a minimum at $A$ and meets the $x$-axis at $B$ and $C$. You are given that $\frac{dy}{dx} = 2x - \frac{2}{x^3}$ and that the curve goes through the point $\left(4, \frac{189}{16}\right)$.

Feb/March 2017

The diagram shows $CXD$ as a semicircle with centre $A$, radius $7\,\text{cm}$ and diameter $CD$. The straight line $YABX$ is perpendicular to $CD$, while arc $CYD$ lies on a circle centred at $B$ with radius $8\,\text{cm}$.

Feb/March 2019

The diagram illustrates a section of the curve with equation $y = \sqrt{x^3 + x^2}$. The shaded area is enclosed by the curve, the $x$-axis and the line $x = 3$.

Feb/March 2019

The diagram depicts a portion of the curve with equation $y = x^2 + 1$. The shaded area enclosed by the curve, the $y$-axis and the line $y = 5$ is rotated through $360^\circ$ about the $y$-axis.

Feb/March 2020

The diagram depicts sector $AOB$, which is a part of a circle centred at $O$ with radius $6\,\text{cm}$ and with angle $AOB = 0.8$ radians. Point $C$ on $OB$ is chosen so that $AC$ is perpendicular to $OB$. Arc $CD$ is part of a circle with centre $O$, and $D$ lies on $OA$.

Feb/March 2020

For the curve given by $y = f(x)$, the derivative is $f'(x) = 2x^{-\frac{1}{3}} - x^{\frac{1}{3}}$. Also, $f(8) = 5$ is known.

Feb/March 2022

The diagram depicts a circle centred at $A$ with radius $5\,\text{cm}$ and another circle centred at $B$ with radius $8\,\text{cm}$. They are tangent at $C$, with $A$, $C$ and $B$ lying on one straight line. The tangent drawn at $D$ on the smaller circle meets the larger circle at $E$ and goes through $B$.

Feb/March 2022

The diagram displays the circle whose equation is $(x - 2)^2 + y^2 = 8$. The chord $AB$ on the circle meets the positive $y$-axis at $A$ and lies parallel to the $x$-axis.

Feb/March 2022

The diagram presents the curve given by $x = y^2 + 1$. The points $A\,(5, 2)$ and $B\,(2, -1)$ are on the curve.

Feb/March 2023

The diagram depicts triangle $ABC$, where angle $B$ is a right angle. $AB$ has length $8\,\text{cm}$ and $BC$ has length $4\,\text{cm}$. Point $D$ lies on $AB$ so that $AD = 5\,\text{cm}$. The sector $DAC$ belongs to a circle with centre $D$.

Feb/March 2023

Work out the exact value of $\int_{3}^{\infty} \frac{2}{x^2}\,dx$.

Feb/March 2024

The diagram presents the curve with equation $y = 2x^{\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1$ for $x > 0$. This curve crosses the $x$-axis at points $A$ and $B$ and has a minimum point $M$.

Feb/March 2024

A curve is defined by $\frac{dy}{dx} = 3\sqrt{4x + 5}$. The points $(1,9)$ and $(5,a)$ are on the curve.

Feb/March 2024

The diagram presents the curve given by $y = 4\sqrt{3x + 4} - 2x - 6$ for $x$ values satisfying $0 \leq x \leq 7$. The tangent drawn to the curve at $P(7, 0)$ intersects the $y$-axis at $Q$. Region $A$ is enclosed by the curve and the two axes. Region $B$ is enclosed by the curve, the line segment $PQ$ and the $y$-axis.

Feb/March 2025

The diagram depicts the curve $y = 6x - x^2$ together with the line $y = 5$.

May/June 2010

The diagram shows a segment of the curve $y = \tfrac{a}{x}$, where $a$ is a positive constant.

May/June 2010

The diagram shows the curve $y = (x-2)^2$ and the line $y + 2x = 7$, which meet at points $A$ and $B$.

May/June 2010

The diagram displays part of the curve $y = x + \frac{4}{x}$, and this curve has a minimum point at $M$. The line $y = 5$ cuts the curve at the points $A$ and $B$.

May/June 2010

Sketch the graph of $y = (x - 2)^2$.

May/June 2011

A television quiz show is broadcast every day. On day 1 the prize money is $\$1000$. If nobody wins it, the prize fund is raised for day 2. This pattern of increase continues each day until someone wins. The television company considered the following two different models for raising the prize money. Model 1: Increase the prize money by $\$1000$ each day. Model 2: Increase the prize money by $10\%$ each day. On each day that the prize money is not won, the television company makes a donation to charity. The donation is $5\%$ of the prize on that day. After $40$ days the prize money has still not been won.

May/June 2011

Calculate $\int \left(x^3 + \frac{1}{x^3}\right)\,dx$.

May/June 2011

The diagram shows a section of the curve $y = 4\sqrt{x} - x$. The curve has its maximum at $M$ and cuts the $x$-axis at $O$ and $A$.

May/June 2011

Differentiate $\frac{2x^3 + 5}{x}$ with respect to $x$.

May/June 2011

The diagram includes the line $y = 1$ together with a section of the curve $y = \frac{2}{\sqrt{x+1}}$.

May/June 2012

The diagram displays the area bounded by the curve $y = \frac{6}{2x - 3}$, the $x$-axis and the vertical lines $x = 2$ and $x = 3$.

May/June 2012

The diagram illustrates a section of the curve $y = -x^2 + 8x - 10$ that goes through the points $A$ and $B$. The curve reaches a maximum at $A$, and the gradient of line $BA$ is $2$.

May/June 2012

The diagram illustrates a section of the curve $x = \frac{8}{y^2} - 2$, which meets the $y$-axis at the point $A$. The point $B(6,1)$ lies on the curve. The shaded region is enclosed by the curve, the $y$-axis and the line $y = 1$.

May/June 2012

A curve is defined by $\frac{dy}{dx} = \frac{6}{x^2}$, and the point $(2, 9)$ lies on it.

May/June 2013

The curve is defined by $\frac{dy}{dx} = \sqrt{2x + 5}$, and it goes through the point $(2, 5)$.

May/June 2013

A line is given by $y = 2x + c$ and a curve by $y = 8 - 2x - x^2$. When the line is tangent to the curve, determine the constant $c$.

May/June 2014

The diagram displays part of the curve $y = 8 - \sqrt{(4 - x)}$ together with the tangent to the curve at $P\,(3, 7)$.

May/June 2014

The curve $y = -x^2 + 12x - 20$ and the line $y = 2x + 1$ are shown in the diagram.

May/June 2014

Consider a curve for which $\frac{dy}{dx} = \frac{12}{\sqrt{4x + a}}$, where $a$ is constant. Point $P(2, 14)$ lies on the curve, and the normal at $P$ has equation $3y + x = 5$.

May/June 2014

The diagram presents a section of the curve $y = \frac{8}{\sqrt{3x + 4}}$. It crosses the $y$-axis at $A(0,4)$. The normal drawn to the curve at $A$ meets the line $x = 4$ at $B$.

May/June 2015

The curve is given by the equation $y = \frac{4}{2x - 1}$.

May/June 2015

The curve is defined by $\frac{dy}{dx} = \sqrt{2x + 1}$, and the point $(4, 7)$ is on it.

May/June 2015

The diagram illustrates a section of the curve $x = \frac{12}{y^2} - 2$. The shaded area is enclosed by the curve, the $y$-axis, and the lines $y = 1$ and $y = 2$.

May/June 2016

The diagram shows the section of the curve $y = \frac{8}{x} + 2x$ for $x > 0$, together with the minimum point $M$.

May/June 2016

The curve satisfies $\frac{dy}{dx} = \frac{8}{(5 - 2x)^2}$. Since it goes through $(2, 7)$,

May/June 2016

The diagram illustrates a section of the curve $y = \sqrt{x^3 + 1}$ together with the point $P(2, 3)$ on the curve.

May/June 2016

The curve satisfies $\frac{dy}{dx}=6x^2+\frac{k}{x^3}$ and it passes through the point $P(1,9)$. The gradient at $P$ is $2$.

May/June 2016

The figure depicts triangle $ABC$, with $AB = 5$ cm, $AC = 4$ cm and $BC = 3$ cm. There are three circles centred at $A$, $B$ and $C$ with radii $3$ cm, $2$ cm and $1$ cm, respectively. The circles are tangent to one another at points $E$, $F$ and $G$, which lie on $AB$, $AC$ and $BC$ respectively.

May/June 2016

The diagram displays part of the curve $y = \frac{4}{5 - 3x}.$

May/June 2017

The diagram displays the straight line $x + y = 5$ crossing the curve $y = \frac{4}{x}$ at the points $A(1, 4)$ and $B(4, 1)$.

May/June 2017

Fig. 1 illustrates a section of the curve $y = x^2 - 1$ together with the line $y = h$, where $h$ is a constant.

May/June 2017

The diagram presents a segment of the curve $y = \frac{x}{2} + \frac{6}{x}$. The line $y = 4$ cuts the curve at the points $P$ and $Q$.

May/June 2018

The diagram displays a section of the curve $y = (x + 1)^2 + (x + 1)^{-1}$ together with the line $x = 1$. Point $A$ is the curve's minimum point.

May/June 2018

The graph of the equation $y = f(x)$ goes through the point $A\,(3,1)$ and intersects the $y$-axis at $B$.

May/June 2018

The diagram depicts triangle $OAB$ with angle $OAB = 90^\circ$ and $OA = 5\ \text{cm}$. Arc $AC$ is part of a circle centred at $O$. Its length is $6\ \text{cm}$, and it cuts $OB$ at $C$.

May/June 2018

The diagram displays a section of the curve $y = \frac{3}{\sqrt{1 + 4x}}$ and a point $P\,(2, 1)$ that lies on the curve. The normal to the curve at $P$ cuts the $x$-axis at $Q$.

May/June 2019

The diagram shows a segment of the curve $y = \sqrt{4x + 1} + \frac{9}{\sqrt{4x + 1}}$ together with the minimum point $M$.

May/June 2019

The figure depicts triangle $ABC$, which is right-angled at $A$. Angle $ABC = \frac{\pi}{5}$ radians, and $AC = 8\,\text{cm}$. Points $D$ and $E$ are located on $BC$ and $BA$ respectively. Sector $ADE$ is part of a circle with centre $A$, and $BDC$ is tangent to arc $DE$ at $D$.

May/June 2019

The diagram depicts a section of the curve $y = \frac{8}{x+2}$ together with the line $2y + x = 8$, which meet at $A$ and $B$. Point $C$ is on the curve, and the tangent at $C$ runs parallel to $AB$.

May/June 2020

The diagram shows a section of the curve $y = \frac{6}{x}$. The points $(1, 6)$ and $(3, 2)$ are on the curve. The shaded area is enclosed by the curve and the lines $y = 2$ and $x = 1$.

May/June 2020

The sketch displays a section of the curve with equation $y = x^3 - 2bx^2 + b^2x$ together with the line $OA$, and $A$ is the maximum point on the curve. The $x$-coordinate of $A$ is $a$, and the curve also has a minimum point at $(b, 0)$, where $a$ and $b$ are positive constants.

May/June 2020

The curve is defined by $\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}$, and the point $(4, 7)$ lies on it.

May/June 2020

The curve is defined by $\frac{dy}{dx} = \frac{3}{x^4} + 32x^3$. It is also stated that the curve goes through $\left(\frac{1}{2}, 4\right)$.

May/June 2021

The diagram displays a section of the curve with equation $y^2 = x - 2$ together with the lines $x = 5$ and $y = 1$. The shaded area bounded by the curve and the lines is turned through $360^\circ$ around the $x$-axis.

May/June 2021

For the curve $y = f(x)$, the derivative is $f'(x) = 6x^2 - \frac{8}{x^2}$. The curve is known to pass through $(2, 7)$.

May/June 2021

The diagram shows a section of the curve with equation $y = x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}$, where $k$ is a positive constant.

May/June 2021

The curve is defined by $y = \sqrt{3x - 2}$, while the straight line is $y = \frac{1}{2}x + 1$. These two graphs meet at points $A$ and $B$.

May/June 2022

For the curve, $\frac{dy}{dx} = 3\sqrt{4x - 7} - 4x^{-\frac{1}{2}}$, and it is known to pass through the point $(4, \frac{5}{2})$.

May/June 2022

The diagram shows the curve given by equation $y = 5x^{\frac{1}{2}}$ and the straight line given by equation $y = 2x + 2$.

May/June 2022

The function $f$ is given by $f(x) = (4x + 2)^{-2}$ for $x > -\tfrac{1}{2}$.

May/June 2022

The diagram displays the curve whose equation is $y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}$. The line $y = 5$ cuts the curve at the points $A (1, 5)$ and $B(16, 5)$.

May/June 2022

The diagram depicts a section of the curve with equation $y = \frac{4}{(2x - 1)^2}$ together with sections of the lines $x = 1$ and $y = 1$. The curve goes through the points $A(1, 4)$ and $B\left(\frac{3}{2}, 1\right)$.

May/June 2023

The curve is given by $\frac{dy}{dx} = \frac{4}{(x - 3)^3}$ for $x > 3$, and it passes through $(4, 5)$.

May/June 2023

The diagram illustrates the curve given by $y = 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}}$ for $x > 0$. The curve crosses the $x$-axis at the points $(0, 0)$ and $(4, 0)$.

May/June 2023

The diagram illustrates the points $A\,(1\tfrac{1}{2}, 5\tfrac{1}{2})$ and $B\,(7\tfrac{1}{2}, 3\tfrac{1}{2})$ on the curve with equation $y = 9x - (2x + 1)^{\tfrac{3}{2}}$.

May/June 2023

The diagram presents part of the curve whose equation is $y = \frac{1}{(5x - 4)^3}$, together with the lines $x = 2.4$ and $y = 1$. The curve meets the line $y = 1$ at the point $(1, 1)$.

May/June 2024

The curve given by $y = 2x - 8x^{\frac{1}{2}}$ reaches a minimum at $A$ and cuts the positive $x$-axis at $B$.

May/June 2024

The diagram presents the curve with equation $y = \sqrt{2x^3 + 10}$.

May/June 2024

The diagram illustrates the curve with equation $y = 5x^2 - 20x$ together with the straight line with equation $y = x - 16$. The $x$-coordinates of the intersection points of the curve and line are 1 and 16.

May/June 2025

The diagram displays the curve given by equation $y = \frac{9}{\sqrt{5x + 4}}$ together with the line $y = 6 - 3x$. The line and the curve meet at point $P$, and $P$ has $y$-coordinate 3.

May/June 2025

If $\int_{1}^{3} \left( \frac{a}{(4x-3)^2} + 2 \right) \, dx = 12$, find the value of the constant $a$.

May/June 2025

A curve is defined by $\frac{dy}{dx} = 12(2x - 5)^2 + 8x$. The curve is known to pass through $(2, 4)$.

May/June 2025

The diagram shows a section of the curve $y = x^2 - \frac{1}{x^2}$. The shaded area is enclosed by the curve, the line $x = 2$ and the $x$-axis.

May/June 2025

Find the value of $\int \left(x + \frac{1}{x}\right)^2 \, dx$.

Oct/Nov 2010

A curve is given by $y = \frac{9}{2 - x}$.

Oct/Nov 2010

The diagram illustrates part of the curve $y = \frac{1}{(3x + 1)^{1/4}}$. The curve intersects the $y$-axis at $A$ and the line $x = 5$ at $B$.

Oct/Nov 2010

The diagram displays sections of the curves $y = 9 - x^3$ and $y = \frac{8}{x^3}$ together with their intersection points $P$ and $Q$. The $x$-coordinates of $P$ and $Q$ are $a$ and $b$ respectively.

Oct/Nov 2010

The diagram represents rhombus $ABCD$. Points $P$ and $Q$ are located on diagonal $AC$ so that $BPD$ is an arc of a circle with centre $C$ and $BQD$ is an arc of a circle with centre $A$. Each side of the rhombus measures $5\text{ cm}$ and angle $BAD = 1.2$ radians.

Oct/Nov 2010

The diagram depicts the curve $y = \sqrt{1 + 2x}$, which crosses the $x$-axis at $A$ and the $y$-axis at $B$. The $y$-coordinate of point $C$ on the curve is $3$.

Oct/Nov 2011

The curve is given by the equation $y = \sqrt{8x - x^2}$.

Oct/Nov 2011

The diagram presents the line $y = x + 1$ together with the curve $y = \sqrt{x + 1}$, which intersect at $(-1, 0)$ and $(0, 1)$.

Oct/Nov 2011

The curve satisfies $\frac{dy}{dx} = -\frac{8}{x^3} - 1$, and the point $(2, 4)$ is on it.

Oct/Nov 2012

The diagram depicts the curve $y^2 = 2x - 1$ together with the straight line $3y = 2x - 1$. These two graphs meet when $x = \frac{1}{2}$ and again when $x = a$, with $a$ being a constant.

Oct/Nov 2012

The diagram depicts part of the curve $y = \frac{9}{2x+3}$, which cuts the $y$-axis at $B(0, 3)$. On the curve, the point $A$ has coordinates $(3, 1)$, and the tangent drawn at $A$ meets the $y$-axis at $C$.

Oct/Nov 2012

A curve is given by $y = \frac{2}{\sqrt{5x - 6}}$.

Oct/Nov 2013

The diagram presents a segment of the curve $y = \frac{8}{x} + 2x$ together with three points $A$, $B$ and $C$ on the curve, whose $x$-coordinates are $1$, $2$ and $5$ respectively.

Oct/Nov 2013

The diagram illustrates the curve $y = \sqrt{x^4 + 4x + 4}$.

Oct/Nov 2013

The curve is described by the equation $y = f(x)$. It is known that $f'(x) = x^{-\frac{3}{2}} + 1$ and that $f(4) = 5$.

Oct/Nov 2013

The diagram presents parts of the curves $y = \sqrt{4x + 1}$ and $y = \frac{1}{2}x^2 + 1$, which intersect at the points $P(0, 1)$ and $Q(2, 3)$. At $Q$, the angle between the tangents to the two curves is $\alpha$.

Oct/Nov 2014

The diagram represents a section of the curve $y = x^2 + 1$.

Oct/Nov 2014

In the diagram, $OADC$ is a sector of a circle with centre $O$ and radius $3\,\text{cm}$. $AB$ and $CB$ are tangents to the circle, and angle $ABC = \frac{1}{3}\pi$ radians. Find, giving your answer in terms of $\sqrt{3}$ and $\pi$,

Oct/Nov 2014

On the diagram, sections of the graphs of $y = x + 2$ and $y = 3\sqrt{x}$ meet at points $A$ and $B$.

Oct/Nov 2014

The diagram shows a section of the curve $y = \sqrt{1 + 4x}$, together with the point $P(6,5)$ on the curve. The line $PQ$ cuts the $x$-axis at $Q(8,0)$.

Oct/Nov 2015

The diagram illustrates part of the curve $y = \sqrt{9 - 2x^2}$. Point $P(2, 1)$ is on the curve, and the normal at $P$ cuts the $x$-axis at $A$ and the $y$-axis at $B$. The shaded area is enclosed by the curve, the $y$-axis and the line $y = 1$.

Oct/Nov 2015

The curve satisfies $\frac{dy}{dx} = \frac{8}{\sqrt{4x + 1}}$, and the point $(2, 5)$ is on it.

Oct/Nov 2016