Further Integration (Edexcel International A Level (IAL) Maths: Pure 4): Exam Questions

3 hours26 questions
1a
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3 marks

Given that u = 3x + 2 show that

(i) du = 3 dx,

(ii) integral3 cos(3x +2) dx = integralcos u du.

1b
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3 marks

Hence find an expression in terms of x for the integral

integral3 cos(3x +2) dx.

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2
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4 marks

Use the substitution u = 4x + 1 to find

integral subscript 2 superscript 6 4 open parentheses 4 x plus 1 close parentheses to the power of 1 half end exponent space d x

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3
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6 marks

Use integration by parts to find an expression for

integral negative 3 x space sin space x space d x.

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4a
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3 marks

Show that

fraction numerator 2 x space minus space 1 over denominator open parentheses x space plus space 1 close parentheses open parentheses x space minus space 2 close parentheses end fraction

can be written in the form

fraction numerator A over denominator x space plus 1 end fraction space plus space fraction numerator B over denominator x space minus 2 end fraction

4b
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4 marks

Hence find

integral fraction numerator 2 x space minus 1 over denominator open parentheses x space plus space 1 close parentheses open parentheses x space minus 2 close parentheses end fraction d x space space space space space x greater than 2

writing your answer as a single logarithm.

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5
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5 marks

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

Graph with a shaded region R under the line y=2x-1, between x=2 and x=4, on a grid with labelled x and y axes from -2 to 10.

(i) For y = 2x + 1, show that y2 = 4x2 + 4x + 1.

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

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6
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6 marks

Use the substitution u = sin x to find the value of

integral subscript 2 superscript pi over 2 end superscript cos squared space 2 x space cos x space d x.

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7
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4 marks

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1
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5 marks

Use a suitable substitution to find

integral-15 sin(5x-2) dx

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2
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7 marks

Use calculus and the substitution u = x + 4 to show that

integral subscript 1 superscript 2 fraction numerator x over denominator x space plus space 4 end fraction d x space equals space 1 plus 4 space I n space 5 over 6

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3
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6 marks

Use integration by parts to find, in terms of e, the exact value of

integral subscript 0 superscript 1(5x - 4)e3x dx

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4a
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3 marks

Show that

fraction numerator 11 over denominator open parentheses 2 x minus 3 close parentheses open parentheses x plus 4 close parentheses end fraction

can be written in the form

fraction numerator A over denominator open parentheses 2 x minus 3 close parentheses end fraction plus fraction numerator B over denominator open parentheses x plus 4 close parentheses end fraction

4b
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4 marks

Hence find

integral fraction numerator 11 over denominator open parentheses 2 x minus 3 close parentheses open parentheses x plus 4 close parentheses end fraction d x

writing your answer as a single logarithm.

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5
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5 marks

The diagram below shows the graph of the curve with equation y = 4 - x2.

A graph of the curve \( y = 4 - x^2 \) shows a shaded region R under the curve above the x-axis. The axes are labelled x and y.

(i) Find the x-coordinates of the points where the graph of y = 4 - x2 intercepts the x-axis.

(ii) The shaded region, R, is to be rotated 360° around the x-axis.

Find the volume of the shape generated.

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1
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5 marks

Use a suitable substitution to find the following

integral 8 x space sin left parenthesis 3 x squared plus 1 right parenthesis space d x

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2
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5 marks

Use the substitution u = 2 + In x to show that

integral fraction numerator 1 over denominator x left parenthesis 2 plus l n space x right parenthesis cubed end fraction space d x space equals fraction numerator negative 1 over denominator 2 left parenthesis ln space x plus space 2 right parenthesis squared end fraction space plus c

where c is the constant of integration.

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3a
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5 marks

Use integration by parts to find

integral left parenthesis 2 x squared minus 1 right parenthesis e to the power of x space d x

3b
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4 marks

Show that

integral I n space x space d x space equals space x space ln space x minus x plus space c

where c is the constant of integration.

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4a
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4 marks

Express

fraction numerator x squared minus 4 x space plus space 7 over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 3 right parenthesis squared end fraction

as partial fractions.

4b
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3 marks

Hence, or otherwise, find

integral fraction numerator x squared minus 4 x plus space 7 over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 3 right parenthesis squared space d x end fraction

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5a
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2 marks

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

Coordinate graph with axes x and y. A line segment connects point O at origin to point (h, r) with vertical and horizontal lines meeting at (h, r).

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

5b
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3 marks

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

Thus use calculus to prove that the general formula for the volume, V, of a cone is

V space equals space 1 third straight pi straight r squared straight h

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

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6
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4 marks

The diagram below shows part of the curve C defined by the equation y space equals space 1 over a x squared, 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.

A graph with shaded region R under a curve from x=1 to x=6 on the x-axis, with axes labelled OX and OY, and the curve's end marked C.

Given that the volume of the solid formed when the region R is rotated 360° about the x-axis is fraction numerator 311 straight pi over denominator 20 end fractioncubic units, find the value of a.

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1a
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3 marks

Find an expression for y given that

y space equals integral 6 x squared e to the power of x cubed end exponent space d x

1b
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3 marks

Integrate

integral left parenthesis 16 minus 32 x right parenthesis space sin left parenthesis 4 x minus 2 right parenthesis squared space d x

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2
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5 marks

Use calculus and the substitution x = cos theta to find the exact value of

integral subscript fraction numerator 1 over denominator square root of 2 end fraction end subscript superscript fraction numerator square root of 3 over denominator 2 end fraction end superscript fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x

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3a
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6 marks

Find

integral x squared space sin space 3 x space d x

3b
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4 marks

Find

integral fraction numerator I n space x over denominator x cubed end fraction space d x

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4
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6 marks

Find the integral

integral fraction numerator 8 x squared minus 8 x minus 1 over denominator left parenthesis 4 x squared minus 1 right parenthesis left parenthesis x minus 2 right parenthesis end fraction d x

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5
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7 marks

Use integration by parts to show that

integral e to the power of x sin x space d x space equals 1 half space e to the power of x left parenthesis sin x minus cos x right parenthesis space plus space C

where c is the constant of integration.

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6a
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3 marks

Sketch the region described by the following inequalities

x ≥ 2 x≤ 6 2y ≤ x + 4 y≥p, where 0<p< 2

6b
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4 marks

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.

6c
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4 marks

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'.

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7
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5 marks

Starting with the equation of a semicircle of radius r, y = square root of r squared minus x squared end root(where r > 0), use calculus to prove that the general formula for the volume, V, of a sphere of radius r is

V space equals 4 over 3 straight pi straight r cubed

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8a
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4 marks

Express fraction numerator 7 y plus 13 over denominator 12 y squared plus 43 y plus 36 end fraction in partial fractions.

8b
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4 marks

Use your answer from part (a) to help find

integral fraction numerator 7 x squared plus 13 over denominator 12 x to the power of 4 plus 43 x squared plus 36 end fraction space d x

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