Basic Calculus Questions - polar integration and roots

1. Aug 25, 2009

perihelion

It's been a while since I studied calculus and basically I have a review sheet for a course I'm taking, but not a graded assignment. So, I was hoping if anyone knew a resource to point me in the right direction with a couple of problems:

$$\int_0^\theta x^a dx$$

Where a is not an element of the set {0, -1} and theta > 0. I'm going over my old calculus text but still at a loss as to how to set this problem up, but maybe I'm making it more difficult than it is.

Also, I had a question involving finding real values of x such that a given function is 0, and justifying my answer. I just did this algebraically, but was wondering if there was another way that I've forgotten using calculus:

$$f(x) = x^2\ +\ 2*x\ +\ 2$$
$$(x+1)*(x+1)+1 = 0$$
$$(x+1)^2 = -1$$
$$x = -1 - i, -1 + i$$

Thanks.

2. Aug 25, 2009

Staff: Mentor

The antiderivative of xa is 1/(a + 1) xa + 1.

3. Aug 25, 2009

perihelion

Thanks, but I was feeling that maybe I was overlooking something. At this point I just stopped.

$$\int_0^\theta x^a dx\ =\ (\theta^(a+1))/(a\ +\ 1)$$

4. Aug 25, 2009

Staff: Mentor

I think you meant this as $\theta^{a + 1}/(a + 1)$
but it didn't quite come out that way.

5. Aug 25, 2009

Staff: Mentor

For your other problem, find the zeroes of f(x) = x2 + 2x + 2, the algebraic (as opposed to calculus) way is one way to go. In fact, calculus is not really appropriate for this type of problem.

f(x) = x2 + 2x + 2 = x2 + 2x + 1 + 1 = (x + 1)2 + 1

The squared term is always >= 0, so adding 1 gives a value that is always >= 1, hence there are no real values for which f(x) = 0.

Looking at this graphically, the function's graph is a parabola that opens up, and whose vertex is at (-1, 1). Since the vertex is the lowest point on the graph, and it is above the x-axis, there are no values of x for which f(x) is less than or equal to zero.