Page 189 - How To Solve Word Problems In Calculus

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Solution
y y
2 2

1 1

x x
1 1 2 2
3
2
The diagram shows the function y = x − 3x + 2x + 1 drawn
with about 20 “imaginary” vertical rectangles of height y and
thickness dx. If we imagine an infinite number of such rectan-
gles extending from x = 0to x = 2, each infinitesimally thin,
the rectangles will “color in” the required area. Since integra-
2

tion is a summation process, ydx will yield the exact area.
0
2 2

2
3
ydx = (x − 3x + 2x + 1) dx
0 0
4 2
x
3
2
= − x + x + x
4 0
= (4 − 8 + 4 + 2) − (0)
= 2
In Example 1 we were given the interval of integration. Often
this interval is determined by the x intercepts of the graph.
These intercepts can be found by setting f (x) = 0 and solving
for x. (It is always advisable to sketch a graph of the region
whose area is to be found.)
EXAMPLE 2
Find the area of the region above the x axis bounded by the
2
function y = 4x − x − 3.
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