Quiggy
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[SOLVED] Improper Integration
Hi, I was working on this problem in my calculus class today and kept getting the answer of 0, while my teacher was saying that the integral diverges. I just want to know if I'm wrong, he's wrong, or something way over my head is going on here and we're both wrong.
The problem was to find:
\int_{-\infty}^{\infty} x^3 dx
I broke this into:
\lim_{b\rightarrow\infty} \int_{-b}^{b} x^3 dx
This in turn worked out to:
\lim_{b\rightarrow\infty} [\frac{1}{4} x^4]_{-b}^{b} = \lim_{b\rightarrow\infty} [(\frac{1}{4} b^4) - (\frac{1}{4} b^4)] = 0
Meanwhile, my teacher broke the original integral into:
\int_{-\infty}^{0} x^3 dx + \int_{0}^{\infty} x^3 dx
This yielded him with indeterminate form \infty - \infty.
So who's right?
Hi, I was working on this problem in my calculus class today and kept getting the answer of 0, while my teacher was saying that the integral diverges. I just want to know if I'm wrong, he's wrong, or something way over my head is going on here and we're both wrong.
The problem was to find:
\int_{-\infty}^{\infty} x^3 dx
I broke this into:
\lim_{b\rightarrow\infty} \int_{-b}^{b} x^3 dx
This in turn worked out to:
\lim_{b\rightarrow\infty} [\frac{1}{4} x^4]_{-b}^{b} = \lim_{b\rightarrow\infty} [(\frac{1}{4} b^4) - (\frac{1}{4} b^4)] = 0
Meanwhile, my teacher broke the original integral into:
\int_{-\infty}^{0} x^3 dx + \int_{0}^{\infty} x^3 dx
This yielded him with indeterminate form \infty - \infty.
So who's right?