Determine the value of the constant y, whenever:
∫(ex - 1)-0.5 dx = pi/6
where ln x denotes the natural logarithm of x.
(In reply to re(2): There are two solutions
Numerical program for the integral from ln(4/3) to ln(8-4abs(sqrt(3))):
pi = ATN(1) * 4
h = -.0000001
y0 = LOG(4 / 3)
yb = y0
yLast = LOG(8 - 4 * ABS(SQR(3)))
x = yb
t = 0
t = t + h / SQR(EXP(x) - 1)
tprev = t
x = x + h
LOOP UNTIL x < yLast
PRINT x - h, tprev, tprev - pi / 6
PRINT x, t, t - pi / 6
6.933646990017295D-02 -.5235990275116614 -1.04719780310996
6.933636990017178D-02 -.5235990275116614 -1.04719780310996
showing the integral to be -pi/6 rather than pi/6.
To be positive the integral would have to be from ln(8-4abs(sqrt(3))) to ln(4/3) rather than the other way around. Any definite integral from a to b is the negative of that integral from b to a.
Posted by Charlie
on 2007-11-28 09:23:19