Evaluate the integral or show that it diverges: Integrand 0, lower, to 3, upper, (1/x)dx.
a. infinity
b. negative infinity
c. 0
d. ln 3
e. none of the above
ln 3 - ln 0
but ln 0 is -oo
so - oo
so ln 3 - ln 0
= + oo
To evaluate the given integral, you can use the fundamental theorem of calculus, which states that if a function F(x) is an antiderivative of f(x), then the integral of f(x) from a to b is equal to F(b) - F(a).
In this case, the integrand is 1/x, which is equivalent to x^(-1). To find the antiderivative of this function, you can use the power rule for integration. According to the power rule, if f(x) = x^n, where n is not equal to -1, then the integral of f(x) with respect to x is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
Applying the power rule to the function x^(-1), you get:
∫(1/x)dx = ln|x| + C
Now, you can find the value of the definite integral from 0 to 3:
∫[0,3] (1/x)dx = ln|x| |_0^3
= ln|3| - ln|0|
Notice that the natural logarithm of 0 is undefined, so you cannot evaluate ln|0|. Since the lower limit of integration is 0, the integral diverges.
Therefore, the correct answer is e) none of the above.