Figure — understand the maths, don't just answer it

DifficultyAll Level 1 Level 2 Level 3 Level 4 Level 5

Indefinite integration (reversing differentiation)

22 questions

LessonNot started

\displaystyle\int x^{1/2} \,\mathrm{d}x

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (8x^3 - 6x^2 + 4) \,\mathrm{d}x

.

●●●●●Level 32 marksStart

Find

\displaystyle\int x^{-2} \,\mathrm{d}x

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (4x^3 - 2x) \,\mathrm{d}x

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (6x^2 + 5) \,\mathrm{d}x

.

●●●●●Level 32 marksStart

The curve

C

has equation

y = \text{f}(x)

The curve

passes through the point $P(-2, 7)$
has a turning point at $P$

Given that

\frac{dy}{dx} = 6x^2 + 2x + k

where

k

is a constant,
(a) show that

k = -20

(2 marks)
(b) Hence find the coordinates of the point where

C

crosses the

y

-axis.

(3 marks)

●●●●●Level 35 marksStart

A conservation officer is monitoring the number of plants of a rare orchid in a meadow after a new footpath brought more visitors to the area.

As a model, the officer proposes that the rate of decline per year of the number of plants is given by

\dfrac{P^{\frac32}}{200}

, where

P

is the number of plants

t

years after the start of the monitoring.
(a) Explain how this model gives rise to the differential equation

\dfrac{dP}{dt} = -\dfrac{P^{\frac32}}{200}

.

(1 mark)

The officer notes that at the start of the monitoring there are

400

plants.
(b) Use the model to determine an expression for

P

in terms of

t

.

(4 marks)

●●●●●Level 35 marksStart

Find

\int \frac{6x^3 - 5x\sqrt{x} + 8}{2x^2}\,\mathrm{d}x

writing your answer in simplest form.

(5 marks)

●●●●●Level 35 marksStart

The constant of integration

17 questions

LessonNot started

A curve passes through

(1, 4)

and has

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x

. Find

y

in terms of

x

.

●●●●●Level 34 marksStart

A curve passes through

(0, 3)

and has

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x - 1

. Find

y

.

●●●●●Level 34 marksStart

A curve passes through

(2, 5)

and has

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2

. Find

y

.

●●●●●Level 34 marksStart

The curve

C

has equation

y = \text{f}(x)

The curve

passes through the point $P(-2, 7)$
has a turning point at $P$

Given that

\frac{dy}{dx} = 6x^2 + 2x + k

where

k

is a constant,
(a) show that

k = -20

(2 marks)
(b) Hence find the coordinates of the point where

C

crosses the

y

-axis.

(3 marks)

●●●●●Level 35 marksStart

Definite integration and area under a curve

60 questions

LessonNot started

In this question you must show detailed reasoning.

The curve

C

has equation

y = \sqrt{2x+1}

,

\quad x \ge -\dfrac{1}{2}

,

and meets the

x

-axis at the point

A

. The tangent to

C

at the point

P

has gradient

\dfrac{1}{3}

.
(a) Find the coordinates of

P

. (3 marks)

The normal to

C

at

P

meets the

x

-axis at the point

Q

. The region

R

is bounded by the arc

AP

of the curve, the line segment

PQ

and the

x

-axis.
(b) Determine the exact area of

R

. (5 marks)

●●●●●Level 38 marksStart

(a) Find

\displaystyle\int 6x\left(x^2 + 1\right)^2 dx

. (3 marks)
(b) Hence evaluate

\displaystyle\int_0^1 6x\left(x^2 + 1\right)^2 dx

. (2 marks)

●●●●●Level 35 marksStart

The table below shows corresponding values of

x

and

y

for

y = \sqrt[3]{x^2 + 5}

The values of

y

are given to 3 decimal places.

\begin{array}{c|cccccc} x & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 \\ \hline y & 1.817 & 1.935 & 2.080 & 2.241 & 2.410 & 2.584 \end{array}

(a) Using the trapezium rule with all the values of

y

in the given table, obtain an estimate for

\int_{1}^{3.5} \sqrt[3]{x^2 + 5}\,\mathrm{d}x

giving your answer to 3 significant figures.

(3 marks)

Using your answer to part (a) and making your method clear, estimate
(b) (i)

\displaystyle\int_{1}^{3.5} \left(4 + \sqrt[3]{x^2 + 5}\right)\mathrm{d}x

(ii)

\displaystyle\int_{1}^{3.5} \sqrt[3]{8x^2 + 40}\,\mathrm{d}x

(3 marks)

●●●●●Level 36 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Show that

\int_0^{\frac{\pi}{3}} 6\sin^2 x\,dx = a\pi + b\sqrt{3}

where

a

and

b

are rational constants to be found.

(4 marks)

●●●●●Level 34 marksStart

Sanjay is investigating the area,

A

, under the graph of

y = \sqrt{1 + x^3}

between

x = 0

and

x = 2

. He draws the graph, together with rectangles of width

\delta x = 0.25

, and varying heights

y

.(a) Use the rectangles in the diagram to show that lower and upper bounds for the area

A

are

3.00

and

3.50

respectively, correct to 3 significant figures.

(1 mark)
(b) Sanjay finds lower and upper bounds for the area

A

, using widths

\delta x

of decreasing size. The results are shown in the table. Where relevant, values are given correct to 3 significant figures.

Width $\delta x$	$0.25$	$0.125$	$0.0625$	$0.03125$
Lower bound for area $A$	$3.00$	$3.12$	$3.18$	$3.21$
Upper bound for area $A$	$3.50$	$3.37$	$3.30$	$3.27$

Use Sanjay's results to estimate the value of

A

correct to 2 significant figures. Give a brief justification for your estimate.

(2 marks)
(c) Write down an expression, in terms of

y

and

\delta x

, for the exact value of the area

A

.

(2 marks)

●●●●●Level 35 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1Figure 1 shows a sketch of part of the curve with equation

y = \dfrac{\sqrt{x}}{x^2 + 1} \qquad x \geqslant 0

The region

R

, shown shaded in Figure 1, is bounded by the curve, the line with equation

x = 1

, the

x

-axis and the line with equation

x = k

, where

k

is a constant and

k > 1

This region is rotated through

360°

about the

x

-axis to form a solid of revolution.

Given that the volume of this solid is

\dfrac{3\pi}{14}

, use algebraic integration to find the exact value of the constant

k

.

(6 marks)

●●●●●Level 36 marksStart

The curve

y = x(4 - x)

and the line

y = x

enclose a finite region.
(a) Find the

x

-coordinates of the points where they meet.
(b) Find the area of the region enclosed between them.

●●●●●Level 35 marksStart

Areas with integration

60 questions

LessonNot started

The curve

C

has equation

y = x^2 - 4x

.
(a) Show that

C

crosses the

x

-axis at

x = 0

and

x = 4

. (2 marks)
(b) The region

R

is enclosed by

C

and the

x

-axis. Explain why

\displaystyle\int_0^4 (x^2 - 4x)\,\mathrm{d}x

is negative, and hence find the area of

R

. (5 marks)

●●●●●Level 37 marksStart

In this question you must show detailed reasoning.

The curve

C

has equation

y = \sqrt{2x+1}

,

\quad x \ge -\dfrac{1}{2}

,

and meets the

x

-axis at the point

A

. The tangent to

C

at the point

P

has gradient

\dfrac{1}{3}

.
(a) Find the coordinates of

P

. (3 marks)

The normal to

C

at

P

meets the

x

-axis at the point

Q

. The region

R

is bounded by the arc

AP

of the curve, the line segment

PQ

and the

x

-axis.
(b) Determine the exact area of

R

. (5 marks)

●●●●●Level 38 marksStart

The diagram shows the curve

y = x(x - 3)

, which lies below the

x

-axis for

0 < x < 3

. Find the area of the shaded region enclosed between the curve and the

x

-axis.

●●●●●Level 33 marksStart

The diagram shows the curve

y = \sqrt{4 - x^2}

for

0 \le x \le 2

, together with the region under it.(a) Use the trapezium rule with

4

strips of equal width to estimate

\displaystyle\int_0^2 \sqrt{4 - x^2}\ \mathrm{d}x

,

giving your answer to 3 significant figures. (4 marks)
(b) Explain, with reference to the shape of the curve, whether the trapezium rule gives an overestimate or an underestimate of the true area. (1 mark)
(c) By considering the shape of the region, write down the exact value of the integral. (1 mark)

●●●●●Level 36 marksStart

The diagram shows the curve

y = x^2

and the line

y = x + 2

. The finite region

R

, shown shaded, is enclosed by the curve and the line. Find the area of

R

.

●●●●●Level 34 marksStart

The diagram shows the curve

y = 9 - x^2

. The shaded region

R

is enclosed between the curve and the

x

-axis. Find the area of

R

.

●●●●●Level 33 marksStart

The diagram shows the curve

y = x^2 - 2x

and the line

y = x

. The finite region

R

, shown shaded, is enclosed by the curve and the line. Find the area of

R

.

●●●●●Level 34 marksStart

The diagram shows the curve

y = \sqrt{x}

and the line

y = x

. The finite region

R

, shown shaded, is enclosed by the curve and the line. Find the area of

R

.

●●●●●Level 34 marksStart

The diagram shows the curve

y = x^2 - 4

. The shaded region

R

is enclosed between the curve and the

x

-axis. Find the area of

R

.

●●●●●Level 33 marksStart

The diagram shows the curve

y = x^3 - x

, which crosses the

x

-axis at

x = -1, 0

and

1

. Find the total area of the two shaded regions enclosed between the curve and the

x

-axis.

●●●●●Level 33 marksStart

Sanjay is investigating the area,

A

, under the graph of

y = \sqrt{1 + x^3}

between

x = 0

and

x = 2

. He draws the graph, together with rectangles of width

\delta x = 0.25

, and varying heights

y

.(a) Use the rectangles in the diagram to show that lower and upper bounds for the area

A

are

3.00

and

3.50

respectively, correct to 3 significant figures.

(1 mark)
(b) Sanjay finds lower and upper bounds for the area

A

, using widths

\delta x

of decreasing size. The results are shown in the table. Where relevant, values are given correct to 3 significant figures.

Width $\delta x$	$0.25$	$0.125$	$0.0625$	$0.03125$
Lower bound for area $A$	$3.00$	$3.12$	$3.18$	$3.21$
Upper bound for area $A$	$3.50$	$3.37$	$3.30$	$3.27$

Use Sanjay's results to estimate the value of

A

correct to 2 significant figures. Give a brief justification for your estimate.

(2 marks)
(c) Write down an expression, in terms of

y

and

\delta x

, for the exact value of the area

A

.

(2 marks)

●●●●●Level 35 marksStart

The curves

y = x^2

and

y = 8 - x^2

enclose a region

R

.(a) Show that the curves intersect where

x = \pm 2

. (2 marks)
(b) Find the area of

R

. (5 marks)

●●●●●Level 37 marksStart

A particle moves in a straight line. At time

t

seconds its velocity is

v = 6t - t^2\ \text{m s}^{-1}

, for

0 \le t \le 4

.
(a) Show that the particle is moving in the positive direction throughout

0 < t \le 4

. (2 marks)
(b) Find the distance travelled in the first

4

seconds. (4 marks)

●●●●●Level 36 marksStart

Integrating standard functions

39 questions

LessonNot started

Find

\displaystyle\int \left(\frac{2}{3x} + 5\sin\tfrac{x}{2}\right) dx

.

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_0^1 e^{2x}\,dx

, giving your answer in terms of

e

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \left(\mathrm{e}^{x} - \mathrm{e}^{-x}\right)^{2} dx

. (5 marks)

●●●●●Level 35 marksStart

Evaluate

\displaystyle\int_{2}^{5} \dfrac{4}{x - 1}\,dx

, giving your answer in the form

k\ln 2

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \dfrac{5}{4 - 2x}\,dx

. (3 marks)

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_1^e \frac{1}{x}\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (e^{3x} + 1)^2\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \left(\cos(2x - 1) + \mathrm{e}^{4 - x}\right) dx

. (4 marks)

●●●●●Level 34 marksStart

Evaluate

\displaystyle\int_{0}^{\pi/6} \cos 3x\,dx

. (3 marks)

●●●●●Level 33 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1Figure 1 shows a sketch of part of the curve with equation

y = \dfrac{5}{\sqrt{2x + 3}} \qquad x > -\dfrac{3}{2}

The finite region

R

, shown shaded in Figure 1, is bounded by the curve, the

y

-axis, the

x

-axis and the line with equation

x = 11

The region

R

is rotated through

360°

about the

x

-axis.

Use calculus to find the exact value of the volume of the solid generated, writing your answer in simplest form.

(5 marks)

●●●●●Level 35 marksStart

Find

\displaystyle\int \left(3e^{-x} + \sec^2 2x\right) dx

.

●●●●●Level 32 marksStart

Integrating with trigonometric identities

24 questions

LessonNot started

Find

\displaystyle\int \sin x\cos x\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (1 + \cos x)^2\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \tan^2 x\,dx

.

●●●●●Level 32 marksStart

Evaluate

\displaystyle\int_{0}^{\pi/4} \tan^{2} x\,dx

, giving your answer exactly. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \cot^{2} x\,dx

. (3 marks)

●●●●●Level 33 marksStart

Show that

\displaystyle\int_{0}^{\pi/2} (\sin x + \cos x)^{2}\,dx = \dfrac{\pi}{2} + 1

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \sin^{2} x\cos^{2} x\,dx

. (5 marks)

●●●●●Level 35 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Show that

\int_0^{\frac{\pi}{3}} 6\sin^2 x\,dx = a\pi + b\sqrt{3}

where

a

and

b

are rational constants to be found.

(4 marks)

●●●●●Level 34 marksStart

Evaluate

\displaystyle\int_0^{\pi/2} \sin^2 x\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int 4\sin^2 3x\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int (\sin x + \cos x)^{2}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \sin 3x\cos 3x\,dx

. (3 marks)

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_0^{\pi/3} \cos^2 x\,dx

, giving your answer in exact form.

●●●●●Level 33 marksStart

Find

\displaystyle\int \cos^{2} 4x\,dx

. (4 marks)

●●●●●Level 34 marksStart

Integration by reversing the chain rule

24 questions

LessonNot started

(a) Find

\displaystyle\int 6x\left(x^2 + 1\right)^2 dx

. (3 marks)
(b) Hence evaluate

\displaystyle\int_0^1 6x\left(x^2 + 1\right)^2 dx

. (2 marks)

●●●●●Level 35 marksStart

Find

\displaystyle\int \mathrm{e}^{2x}\left(\mathrm{e}^{2x} + 1\right)^{3}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \dfrac{4x - 6}{x^{2} - 3x + 5}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \sec^{2} x\tan^{3} x\,dx

. (3 marks)

●●●●●Level 33 marksStart

Find

\displaystyle\int \cos x\,e^{\sin x}\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int x\sqrt{x^2 + 5}\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int \frac{2x + 3}{x^2 + 3x + 1}\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int \sin 2x\,(1 + \cos 2x)^{2}\,dx

. (5 marks)

●●●●●Level 35 marksStart

Find

\displaystyle\int x\,e^{x^2}\,dx

.

●●●●●Level 32 marksStart

Evaluate

\displaystyle\int_0^1 \frac{x}{x^2 + 1}\,dx

, giving your answer in terms of

\ln

.

●●●●●Level 33 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1Figure 1 shows a sketch of part of the curve with equation

y = \dfrac{\sqrt{x}}{x^2 + 1} \qquad x \geqslant 0

The region

R

, shown shaded in Figure 1, is bounded by the curve, the line with equation

x = 1

, the

x

-axis and the line with equation

x = k

, where

k

is a constant and

k > 1

This region is rotated through

360°

about the

x

-axis to form a solid of revolution.

Given that the volume of this solid is

\dfrac{3\pi}{14}

, use algebraic integration to find the exact value of the constant

k

.

(6 marks)

●●●●●Level 36 marksStart

(a) Find

\dfrac{d}{dx}\left(5 - 2x^2\right)^4

(2 marks)
(b) Hence show that

\int_0^1 x\left(5 - 2x^2\right)^3 dx = R

where

R

is an integer to be found.

(Solutions relying on calculator technology are not acceptable.)

(3 marks)

●●●●●Level 35 marksStart

Find

\displaystyle\int \sin x\cos^4 x\,dx

.

●●●●●Level 32 marksStart

Find

\displaystyle\int \dfrac{x}{(x^{2} + 4)^{2}}\,dx

. (3 marks)

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_{0}^{1} x(x^{2} + 3)^{3}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Integration by substitution

36 questions

LessonNot started

Using the substitution

u = 2x + 1

, find

\displaystyle\int x\sqrt{2x + 1}\,dx

.

●●●●●Level 34 marksStart

Using the substitution

u = x + 3

, find

\displaystyle\int x\sqrt{x + 3}\,dx

. (5 marks)

●●●●●Level 35 marksStart

Using the substitution

u = \ln x

, find

\displaystyle\int \dfrac{(\ln x)^{3}}{x}\,dx

. (3 marks)

●●●●●Level 33 marksStart

Using the substitution

u = \sin x

, evaluate

\displaystyle\int_{0}^{\pi/2} \cos x\sin^{4} x\,dx

. (4 marks)

●●●●●Level 34 marksStart

Using the substitution

u = x^2 + 1

, evaluate

\displaystyle\int_0^1 \frac{x}{(x^2 + 1)^2}\,dx

.

●●●●●Level 34 marksStart

Using a suitable substitution, find

\displaystyle\int x^2(x^3 + 1)^5\,dx

.

●●●●●Level 33 marksStart

Integration by parts

32 questions

LessonNot started

(a) Find

\displaystyle\int x\cos x; dx

. (4 marks)
(b) Hence find the exact value of

\displaystyle\int_0^{\pi/3} x\cos x; dx

. (3 marks)
(c) Explain why your answer to part (b) gives the area of the region bounded by the curve

y = x\cos x

, the

x

-axis and the line

x = \dfrac{\pi}{3}

. (2 marks)

●●●●●Level 39 marksStart

Find

\displaystyle\int \ln x\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int x\sin 2x\,dx

.

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_0^1 x\,e^{2x}\,dx

, giving your answer in terms of

e

.

●●●●●Level 34 marksStart

Find

\displaystyle\int x^2 e^x\,dx

.

●●●●●Level 34 marksStart

Find

\displaystyle\int x^2\ln x\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int x^2\cos x\,dx

.

●●●●●Level 34 marksStart

Integrating with partial fractions

45 questions

LessonNot started

Find

\displaystyle\int \frac{5}{(2x + 1)(x - 2)}\,dx

.

●●●●●Level 33 marksStart

Evaluate

\displaystyle\int_2^4 \frac{4}{(x - 1)(x + 1)}\,dx

, giving your answer as a single logarithm.

●●●●●Level 34 marksStart

Find

\displaystyle\int \dfrac{6}{(x + 1)(x + 4)}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \dfrac{x^{2} + 3x + 3}{(x + 1)(x + 2)}\,dx

. (5 marks)

●●●●●Level 35 marksStart

Given that

\dfrac{2}{x(x - 1)^2} = \dfrac{2}{x} - \dfrac{2}{x - 1} + \dfrac{2}{(x - 1)^2}

, find

\displaystyle\int \frac{2}{x(x - 1)^2}\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \dfrac{2x + 3}{(x + 1)(x + 2)}\,dx

. (4 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \frac{3x + 5}{(x + 1)(x + 3)}\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \frac{x + 4}{x^2 - x - 2}\,dx

.

●●●●●Level 34 marksStart

Find

\displaystyle\int \dfrac{3x + 4}{(x + 1)^{2}}\,dx

. (5 marks)

●●●●●Level 35 marksStart

Find

\displaystyle\int \dfrac{x + 6}{x(x + 2)(x - 3)}\,dx

. (6 marks)

●●●●●Level 36 marksStart

Find

\displaystyle\int \frac{4}{(x - 1)(x + 1)}\,dx

.

●●●●●Level 33 marksStart

Find

\displaystyle\int \frac{x + 7}{(x + 2)(x - 3)}\,dx

.

●●●●●Level 33 marksStart

\text{f}(x) = \frac{14}{(x - 2)(x + 5)}\qquad x > 2

(a) Express

\text{f}(x)

in partial fractions.

(2 marks)
(b) Hence find

\displaystyle\int \text{f}(x)\,dx

(2 marks)

●●●●●Level 34 marksStart

Find

\displaystyle\int \frac{x^2}{x^2 - 1}\,dx

.

●●●●●Level 34 marksStart

Areas under parametric curves

28 questions

LessonNot started

The diagram shows the curve with parametric equations

x = t^2,\ y = t

. The shaded region is bounded by the curve, the

x

-axis and the line

x = 9

. Find its area.

●●●●●Level 33 marksStart

A curve has parametric equations

x = t + 1

,

y = 4 - t^{2}

. The curve meets the

x

-axis where

t = -2

and

t = 2

.

Find the area of the region enclosed by the curve and the

x

-axis. (5 marks)

●●●●●Level 35 marksStart

A curve has parametric equations

x = 2t,\ y = t^2 + 1

. The shaded region is bounded by the curve, the

x

-axis and the lines

x = 0

and

x = 4

. Find its area.

●●●●●Level 33 marksStart

A curve has parametric equations

x = 3t

,

y = t^{2}

. Find the area of the region bounded by the curve, the

x

-axis, and the lines corresponding to

t = 0

and

t = 2

. (4 marks)

●●●●●Level 34 marksStart

A curve has parametric equations

x = t^{2}

,

y = 2t + 1

, for

t \geqslant 0

. Find the area of the region bounded by the curve, the

x

-axis, and the lines corresponding to

t = 1

and

t = 3

. (5 marks)

●●●●●Level 35 marksStart

A curve has parametric equations

x = t

,

y = t^{2} - 4

. Find the area of the finite region bounded by the curve and the

x

-axis between

t = 0

and

t = 2

. (4 marks)

●●●●●Level 34 marksStart

Differential equations

56 questions

LessonNot started

Find the general solution of

\dfrac{dy}{dx} = 3x^2 y

.

●●●●●Level 33 marksStart

Solve

\dfrac{dy}{dx} = \dfrac{2x}{y}

given that

y = 3

when

x = 0

.

●●●●●Level 34 marksStart

(a) Find the general solution of the differential equation

(x^2 + 4)\dfrac{dy}{dx} = 4xy

.
(b) Find the particular solution for which

y = 1

when

x = 0

.

●●●●●Level 36 marksStart

Find the particular solution of the differential equation

\dfrac{dy}{dx} = x\,e^{x - y}

for which

y = \ln 2

when

x = 0

, giving your answer in the form

y = f(x)

.

●●●●●Level 35 marksStart

Find the particular solution of

\dfrac{dy}{dx} = (2x + 1)e^{-y}

for which

y = 0

when

x = 0

, giving your answer in the form

y = f(x)

.

●●●●●Level 34 marksStart

(a) Find the general solution of

\dfrac{dy}{dx} = y\tan x

for

0 < x < \dfrac{\pi}{2}

.
(b) Find the particular solution for which

y = 2

when

x = 0

.

●●●●●Level 35 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that

y = 0

when

x = 0

, solve the differential equation

\frac{dy}{dx} = e^{-y}\cos x

giving your answer in the form

y = \text{f}(x)

.

(4 marks)

●●●●●Level 34 marksStart

Find the general solution of

\dfrac{dy}{dx} = \dfrac{y}{x}

for

x > 0

, giving

y

explicitly.

●●●●●Level 33 marksStart

An empty garden pond has vertical sides and a horizontal base of area

4

square metres.

The pond is filled with water from a hose at a rate of

0.06

cubic metres per minute.

At time

t

minutes after the hose is switched on, the depth of water in the pond is

h

metres and water soaks away through the base of the pond at a rate of

0.02h

cubic metres per minute.

At time

t

minutes after the hose is switched on, the volume of water in the pond is

V

cubic metres.

Show that

\frac{dV}{dt} = \frac{12 - V}{200}

(4 marks)

●●●●●Level 34 marksStart

A population

P

satisfies

\dfrac{dP}{dt} = kP

. It is

1000

when

t = 0

and

1500

when

t = 2

. Find

P

in terms of

t

.

●●●●●Level 34 marksStart

A conservation officer is monitoring the number of plants of a rare orchid in a meadow after a new footpath brought more visitors to the area.

As a model, the officer proposes that the rate of decline per year of the number of plants is given by

\dfrac{P^{\frac32}}{200}

, where

P

is the number of plants

t

years after the start of the monitoring.
(a) Explain how this model gives rise to the differential equation

\dfrac{dP}{dt} = -\dfrac{P^{\frac32}}{200}

.

(1 mark)

The officer notes that at the start of the monitoring there are

400

plants.
(b) Use the model to determine an expression for

P

in terms of

t

.

(4 marks)

●●●●●Level 35 marksStart

Solve

\dfrac{dy}{dx} = \dfrac{\cos x}{y}

given that

y = 2

when

x = 0

.

●●●●●Level 34 marksStart

Solve

\dfrac{dy}{dx} = x y^2

given that

y = 1

when

x = 0

, giving

y

explicitly.

●●●●●Level 34 marksStart

A hot drink cools so that

\dfrac{d\theta}{dt} = -k(\theta - 20)

, where

\theta

is its temperature in °C. It is

80

°C at

t = 0

and

50

°C at

t = 5

minutes. Find

\theta

in terms of

t

.

●●●●●Level 35 marksStart