With the ordinary method we would have had to apply the Parts Formula four
times to perform this integration.
E. APPLICATIONS OF ANTIDERIVATIVES; DIFFERENTIAL
EQUATIONS
The following examples show how we use given conditions to determine
constants of integration.
Example 48 __
Find f (x) if f ′(x) = 3x^2 and f (1) = 6.
.SOLUTION:
Since f (1) = 6, 1^3 + C must equal 6; so C must equal 6 − 1 or 5, and f (x) = x^3 +
5.
Example 49 __
Find a curve whose slope at each point (x, y) equals the reciprocal of the x-value
if the curve contains the point (e, −3).
SOLUTION: We are given that and that y = −3 when x = e. This equation
is also solved by integration. Since.
Thus, y = ln x + C. We now use the given condition, by substituting the point (e,
−3), to determine C. Since −3 = ln e + C, we have −3 = 1 + C, and C = −4. Then,
the solution of the given equation subject to the given condition is
y = ln x − 4.(^1) This method was described by K. W. Folley in Vol. 54 (1947) of the American Mathematical Monthly and