Indefinite Integrals Exercises

Answer

This function is badly behaved at the upper limit of integration x = 0. We write the improper integral as a limit, work out the integral, then evaluate the limit.

This integral diverges, since  approaches ∞ as b approaches 0. That means the area of this region is infinite.

Answer

This function is badly behaved at x = 0, which is the lower limit of integration. We write out the improper integral as the limit it is, then evaluate:

As b approaches 0, the quantity b^0.6 also approaches 0. So the integral converges to .

Answer

This function is badly behaved at x = 1, which is the lower limit of integration. We write the improper integral as the appropriate limit and work out the integral:

As b approaches 1 the quantity (b – 1)^1/2 approaches 0. The integral converges to 2.

Answer

This integral is badly behaved at x = 0. We worked out the antiderivative of  in another problem.

Since the quantity p – 1 is positive, as b approaches 0 the quantity b^{p – 1} will also approach 0. Then the fraction

will approach -∞. The limit does not exist, so the integral

diverges.

Answer

This integral is also badly behaved at x = 0. Most of the work is the same as in the previous problem. We don't need to change anything until we get to this step:

At this point, since p < 1, the exponent p – 1 is negative. As b approaches 0 the quantity b^{p – 1} will approach ∞, and so the term

will approach 0. The limit converges to

Since p < 1 the quantity  is positive, which is reassuring since the area between  and the x-axis on (0, 1] is all above the x-axis.

Answer

False. Take the graph of  on the interval [-1,0). This graph is the graph of  reflected over the y-axis. Then  does not exist and

converges.

Answer

False. If we take the graph of  on the interval [-1,0), then  does not exist and

diverges.

Answer

This function is definitely weird at 0. We break up the integral into two pieces, splitting the interval at x = 0:

Let's integrate the one with non-negative limits of integration first, since this integral looks more likely to diverge and save us work.

As b approaches 0 the quantity  approaches ∞, which means e^1/b approaches ∞ also. This integral diverges, which means the original integral diverges. We don't have to bother with the other integral.

Answer

This function is weird at x = 1. So we split the integral at x = 1.

Let's start with the right-most integral. When x is in the interval (1, 2] the quantity x – 1 is positive, so we can drop the absolute value signs in the denominator.

As b approaches 1, the quantity b – 1 approaches 0 and so the quantity 2((b) – 1)^1/2 approaches 0 also. This means we have a convergent integral:

Now for the other integral. Since x – 1 is negative when 0 < x < 1, we can get rid of the absolute value signs in the denominator by writing

|x – 1| = -(x – 1) = 1 – x.

We have to use substitution to find the antiderivative, since x has a coefficient of -1.

As b approaches 1, the quantity 2(1 – (b))^1/2 approaches 0, so we're left with

Putting everything together, we see that our original integral converges: