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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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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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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