Unit 2 Revision Exercise 3

Solutions – Integration

1.

= passes through (4,) ( x > 0 )

= c = 1

So

2.

= passes through (8,8)

8 = c = -2

So

3. y = -3 + 4x – x²

= (3-x)(x-1) so A(1,0) and B(3,0) use the values of A and B as the limits.

Integrate in two parts since one part lies beneath the x – axis

Shaded area = +

= +

= ( -3 + 2 - ) + [( -9 + 18 – 9 ) - ( -3 + 2 - )]

= (negative since the area lies beneath the x – axis – so ignore sign) +

total shaded area is + = square units

4. y = x³ – 6x² + 12x – 7

y(1) = 1 – 6 + 12 – 7 = 0 (x-1) is a factor by the factor theorem

1 1 -6 12 -7

1 -5 7

1 -5 7 0

Factors are (x-1)(x²-5x+7)

examining the quadratic factor: b²-4ac<0 no real roots

So y has only one real root at x = 1

To find the gradient of the tangent at x = 3 first find

= 3x² - 12x + 12

at x = 3 = 27 – 36 + 12 = 3 So gradient of tangent at x = 3 is 3

To find the y coordinate substitute x = 3 into y = x³ – 6x² + 12x – 7 y = 2

gradient 3 point (3,2) equation

y – b = m(x-a)

y – 2 = 3(x-3)

y = 3x – 7 equation of tangent at x = 3

To find the area bounded by these two curves

y = x³ – 6x² + 12x – 7 Let this upper curve be f(x)

y = 3x – 7 Let this lower curve be g(x)

To find the limits let x³ – 6x² + 12x – 7 = 3x – 7

x³ – 6x² + 9x = 0

x(x² – 6x + 9) =0

x(x-3)² = 0

limits are x = 0 and x = 3

Shaded area =

=

=

=

=

= = square units