Could someone please help me solve the following questions re the graph?

Method 1 , Calculus method
f(x) has two turning points A( 5, 2 ) , B( 17 , - 1 )
slope of AB = ( - 1 - 2 )/( 17 - 5 ) = - 3/12 = - 1/4
So the eq. of AB is y - 2 = - 1/4( x - 5 ) , that is
y = - (1/4)x + 13/4

Hence,
f(x) = 2 , when 0 ≦ x < 5
f(x) = - (1/4)x + 13/4 , when 5 ≦ x ≦ 17
f(x) = - 1 , when 17 < x ≦ 24

∫ f(x) dx , from x = 0 to x = 5
= ∫ 2 dx
= [ 2x ] , from x = 0 to x = 5
= 2*5 - 2*0
= 10

∫ f(x) dx , from x = 5 to x = 17
= ∫ [ - (1/4)x + 13/4 ] dx
= [ - (1/8)x^2 + (13/4)x ] , from x = 5 to x = 17
= [ - (1/8)17^2 + (13/4)17 ] - [ - (1/8)5^2 + (13/4)5 ]
= - 289/8 + 221/4 + 25/8 - 65/4
= 6

∫ f(x) dx , from x = 17 to x = 24
= ∫ (-1) dx
= [ - x ] , from x = 17 to x = 24
= -24 - (-17)
= - 7

∫ [0,24] f(x) dx
= ∫ [0,5] f(x) dx + ∫ [5,17] f(x) dx + ∫ [17,24] f(x) dx
= 10 + 6 - 7
= 9 ..... Ans

----------------------------------------------------------------------------------------

Method 2 , geometry method
Let
A1 = the trapezoid area for f(x) ≧ 0
A2 = the trapezoid area for f(x) < 0
Then
A1 = (5+13) * 2 / 2 = 18
A2 = (11+7) * 1 / 2 = 9

∫ [0,24] f(x) dx
= A1 - A2
= 18 - 9
= 9