Indefinite Integrals Exercises

Answer

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

diverges.

Answer

Since -1 < sin x < 1, we know

Since

diverges, the integral

also diverges. Since the original integrand

 is greater than , the original integral

also diverges.

Answer

If we make the denominator smaller we make the fraction bigger, so

Since we know

converges, the integral with smaller integrand,

also converges.

Answer

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

diverges. Since

is a constant multiple of , this means

diverges also. Finally, we conclude

diverges.

Answer

For x > 1 we know that sin x < x². This means

0 <

Since

converges, so does

and therefore so does

Answer

While we can't integrate this integral exactly, for x > 1 we know that x² > x. This means

0 <

Since

converges, so does

Answer

Since ln x < x, we know . Since

diverges, so does

.