L&T, 18.1-18
We have already discussed how an integral corresponds to an area.
Example 6.1:
Evaluate the area under from to .
Solution:
which is , see Fig. 6.1 .
Example 6.2:
Find the area of the region bounded by and , for ranging from to , see Fig. 6.2 .
Solution:
From the graph we see that is above , so that
Here we have made the optional choice to combine the two integrands before evaluation of the integral.
L&T, 18.1-18
To find area beneath the curve between and , we divide the area into strips as shown in Fig. 6.3 . Let the thickness of strip at be . The height at is , and therefore the area of the strip is . Now sum up all strips from to . The areas is
In the limit that becomes infinitesimal (i.e., approaches zero), we replace by , the by and so
(6.1) |