Page 192 - How To Solve Word Problems In Calculus

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We integrate separately on the three intervals [0, 2], [2, 3], and
[3, 4].

2 x 3 5x 2
2
2
I 1 = (x − 5x + 6) dx = − + 6x
0 3 2 0
8 20 14

= − + 12 − (0) =
3 2 3
3 x 3 5x 2
3
2
I 2 = (x − 5x + 6) dx = − + 6x
2 3 2 2

45 8 20
= 9 − + 18 − − + 12
2 3 2
9 14 1
= − =−
2 3 6
4 x 3 5x 2
4
2
I 3 = (x − 5x + 6) dx = − + 6x
3 3 2 3

64 45
= − 40 + 24 − 9 − + 18
3 2
16 9 5
= − =
3 2 6
To obtain the required area, we add the absolute values of
I 1 , I 2 , and I 3 .
14 1 5 17
Area =|I 1 |+|I 2 |+|I 3 |= + + =
3 6 6 3
To determine the area bounded by two curves, y = f (x)
and y = g(x), we must first determine their points of intersec-
tion. This may be done by solving the equation f (x) = g(x). If
the curves intersect at only two locations, say x = a and x = b,
and f (x) lies above g(x), i.e., f (x) ≥ g(x) for x ε [a, b], the area
will be
b

A = [ f (x) − g(x)] dx
a

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