Determine if the integral converges or diverges. What integral are you using for comparison in each case?
The integrand is a constant multiple of the function . We know that
diverges because 0.2 < 1,
and we know that
diverges because changing the lower limit of integration doesn't change whether the integral converges or diverges.
This means the integral
Since -1 < sin x < 1, we know
diverges, the integral
also diverges. Since the original integrand
is greater than , the original integral
If we make the denominator smaller we make the fraction bigger, so
Since we know
converges, the integral with smaller integrand,
for x > 1, we know sin x < x and )
Since the integrand looks sort of like , we're going to guess this integral diverges. To show divergence for certain, we need to find a function that is less than
whose integral diverges. Following the first hint, let's make the denominator bigger (and the function smaller) by replacing sin x with x:
Now following the second hint, make the denominator bigger again (and the function smaller again) by replacing square root with x:
We know that
is a constant multiple of , this means
diverges also. Finally, we conclude
For x > 1 we know that sin x < x2. This means
converges, so does
and therefore so does
While we can't integrate this integral exactly, for x > 1 we know that x2 > x. This means
Since ln x < x, we know . Since
diverges, so does
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