Page 208 - How To Solve Word Problems In Calculus

P1: FYN/FZQ P2: FXS
PB056-07 Macros-FMT June 1, 2001 13:49 Char Count= 0

Since the graph falls below the x axis, we must integrate separately
over two intervals.

1 3

3
2
3
2
I 1 = (x − 4x + 3x) dx I 2 = (x − 4x + 3x) dx
0 1
4 3 2 1 4 3 2 3
x 4x 3x x 4x 3x
= − + = − +
4 3 2 4 3 2
0 1

1 4 3 81 108 27
= − + − 0 = − +
4 3 2 4 3 2

5 1 4 3
= − − +
12 4 3 2
9 5
=− −
4 12
8
=−
3
I 2 is negative since the curve falls below the x axis between x = 1
and x = 2. The total area is computed by adding the absolute values
of the integrals.

5 8 37
A =|I 1 |+|I 2 |= + =
12 3 12

3. (a)

(x, ) 8 y = 8
8 8

y = x 3

(x,y)

1 1 2 2

195