does the integral dx over x to the third converge or diverge with bounds of 1 and infinity?

Answer 1

We have to compute Integral (1 to oo) dx/x^3. = Integral (1 to oo) x^(-3) dx. We know the primitive of x^(-3) is 1/(-3 +1) x^(-3 +1) = -1/2 x^(-2). So, we have [-1/2 x^(-2)] (1 to oo). Since the exponent of x is negative, this function goes to zer when x goes to oo. And our integral is - (-1/2) 1^(-2) = 1/2, which, therefore, converges.

By a similar reasoning we can come to a general result: if 0, p <1, then Integral (1 to oo) (1/x^p) dx diverges. If p >1, the integral converges to 1/(p -1). In your case, p =3

Answer 2

Integral of (1/x^3)dx is -1/(2x^2)

Plug in the values.

It converges

Answer 3

S x^-3 dx = (-1/2)x^-2
lim (n => infinity) (-1/2)(n^-2) - (-1/2)(1)^-2
1/2

the integral converges