Integration

Please choose a question to attempt from the following:

1 2 3 4 5

y = x2 - 8x + 18

x = 3 x = k

Show that the shaded area is given by

1/3k3 – 4k2 + 18k - 27

INTEGRATION : Question 1

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The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.

The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.

y = x2 - 8x + 18

x = 3 x = k

Show that the shaded area is given by

1/3k3 – 4k2 + 18k - 27

INTEGRATION : Question 1

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Area = (x2 - 8x + 18) dx3

k

= 1/3k3 – 4k2 + 18k – 27 as required.

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Question 1

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The diagram shows the

curve y = x2 - 8x + 18 and the

lines x = 3 and x = k.

Show that the shaded area is

given by 1/3k3 – 4k2 + 18k - 27

Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

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Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

• Learn result

can be used to find the

enclosed area shown:

( )b

a

f x dx

a b

f(x)

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Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

• Learn result for integration:

“Add 1 to the power and divide by the new power.”

INTEGRATION : Question 2

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Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes

through the point (2,15) then find the equation of the curve

y = f(x).

INTEGRATION : Question 2

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Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes

through the point (2,15) then find the equation of the curve

y = f(x).

Equation of curve is y = 4x3 – 3x2 - 5

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Question 2

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Given that dy/dx = 12x2 – 6x and

the curve y = f(x) passes through

the point (2,15) then find the

equation of the curve y = f(x).

dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C

Substituting (2,15) into y = 4x3 – 3x2 + C

We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5

Equation of curve is

y = 4x3 – 3x2 - 5

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dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C

Substituting (2,15) into y = 4x3 – 3x2 + C

We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5

Equation of curve is

y = 4x3 – 3x2 - 5

• Learn the result that integration undoes differentiation:

i.e. given

= f(x) y = f(x) dx dy

dx

• Learn result for integration:

“Add 1 to the power and divide by the new power”.

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dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C

Substituting (2,15) into y = 4x3 – 3x2 + C

We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5

Equation of curve is

y = 4x3 – 3x2 - 5

• Do not forget the constant of integration!!!

INTEGRATION : Question 3

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Find x2 - 4 2xx

dx

INTEGRATION : Question 3

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Find x2 - 4 2xx

dx

= xx + 4 + C 3 x

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Find x2 - 4 2xx

dxx2 - 4 2xx

dx= x2 - 4

2x3/2 2x3/2dx

= 1/2x1/2 - 2x-3/2 dx

= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C

= 1/3x3/2 + 4x-1/2 + C

= xx + 4 + C 3 x

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x2 - 4 2xx

dx= x2 - 4

2x3/2 2x3/2dx

= 1/2x1/2 - 2x-3/2 dx

= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C

= 1/3x3/2 + 4x-1/2 + C

= xx + 4 + C 3 x

• Prepare expression by:

1 Dividing out the fraction. 2 Applying the laws of indices.

• Learn result for integration:

Add 1 to the power and divide by the new power.

• Do not forget the constant of integration.

INTEGRATION : Question 4

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1

2

( )Evaluate x2 - 2 2 dx x

INTEGRATION : Question 4

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1

2

( )Evaluate x2 - 2 2 dx x

= 21/5

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Question 4

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1

2

( )Evaluate x2 - 2 2 dx x

= 21/5

1

2

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

2

( )= x4 - 4x + 4x-2 dx1

2

[ ]= x5 - 4x2 + 4x-1 5 2 -1 1

2

= x5 - 2x2 - 4 5 x[ ]2

1

= (32/5 - 8 - 2) - (1/5 - 2 - 4)

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Next Comment= 21/5

1

2

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

2

( )= x4 - 4x + 4x-2 dx1

2

[ ]= x5 - 4x2 + 4x-1 5 2 -1 1

2

= x5 - 2x2 - 4 5 x[ ]2

1

= (32/5 - 8 - 2) - (1/5 - 2 - 4)

• Prepare expression by:

1 Expanding the bracket2 Applying the laws of indices.

• Learn result for integration:

“Add 1 to the power and divide by the new power”.

• When applying limits show substitution clearly.

(a) Find the coordinates of A and B.

(b)Hence find the shaded area between the

curves.

y = -x2 + 8x - 10

y = x

A

B

INTEGRATION : Question 5

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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.

(a) Find the coordinates of A and B.

(b)Hence find the shaded area between the

curves.

y = -x2 + 8x - 10

y = x

A

B

INTEGRATION : Question 5

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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.

A is (2,2) and B is (5,5) .

= 41/2units2

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Question 5

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The diagram shows the parabola

y = -x2 + 8x - 10 and the line y = x.

They meet at the points A and B.

(a) Find the coordinates of A and B.

(b) Hence find the shaded area

between the curves.

(a) Line & curve meet when

y = x and y = -x2 + 8x - 10 .

So x = -x2 + 8x - 10

or x2 - 7x + 10 = 0

ie (x – 2)(x – 5) = 0

ie x = 2 or x = 5

Since points lie on y = x then

A is (2,2) and B is (5,5) .

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Question 5

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The diagram shows the parabola

y = -x2 + 8x - 10 and the line y = x.

They meet at the points A and B.

(a) Find the coordinates of A and B.

(b) Hence find the shaded area

between the curves.

A is (2,2) and B is (5,5) .

(b) Curve is above line between limits so

Shaded area = (-x2 + 8x – 10 - x) dx2

5

= (-x2 + 7x – 10) dx2

5

= -x3 + 7x2 - 10x3 2[ ] 5

2

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

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(a) Line & curve meet when

y = x and y = -x2 + 8x - 10 .

So x = -x2 + 8x - 10

or x2 - 7x + 10 = 0

ie (x – 2)(x – 5) = 0

ie x = 2 or x = 5

Since points lie on y = x then

A is (2,2) and B is (5,5) .

• At intersection of line and curve

yy11 = y = y22

Terms to the left, simplify and Terms to the left, simplify and factorise.factorise.

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(b) Curve is above line between limits so

Shaded area = (-x2 + 8x – 10 - x) dx2

5

= (-x2 + 7x – 10) dx2

5

= -x3 + 7x2 - 10x3 2[ ] 5

2

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

• Learn result

can be used to find the

enclosed area shown:

b

2 1

a

Area = (y y )dx

a b

upper curve

y1area

y2

lower curve