Quiz 7 Solution

1. Which of these is (are) improper?

See the definition of an improper integral, Thomas/Finney p. 595. 'a' is clearly improper. So is 'b' since the integrand goes to infinity between -1 and 1 (That is, as x approaches 0, the integrand goes to infinity). Again, so is 'd' since it's integrand goes to infinity at the endpoints.

2.

$$I=\int_0^3 x^r\,dx = \left\{ \begin{array}{ll} \left. \frac{x... ... \ln x \right\vert _0^3 &\mbox{if $r=-1$ } \end{array} \right.$$

case 1: $r\ne -1$
Then

$$I = \frac{3^{r+1}}{r+1} - \frac{0^{r+1}}{r+1}$$

The first term is easy to calculate since $r\ne -1$. If r>-1, then r+1>0, and so the second term is zero. But if r<-1, we have r+1<0 and so the second term is undefined (but goes to infinity if we take the limit). So, in this case, the integral converges if and only if r>-1.

case 2: r=-1
Then

$$I = \ln 3 - \ln 0$$

Again, the first term is ok. But the second term is undefined, and if we take the limit, it goes to $-\infty$. Therefore, the integral does not converge in this case.

So, by these 2 cases, we see that the integral converges if and only if r>-1.

You are visitor