Question

another prob. impr. integral.

Answer 1

\[\int\limits_{1}^{4}\frac{x}{x^2-16}dx\]I will do it now.

Answer 2

\[\lim_{a \rightarrow 4^-} \int\limits_{1}^{a}\frac{x}{x^2-16}dx\]\[u=x^2-16\]\[\lim_{a \rightarrow 4^-} \frac{1}{2}\int\limits_{1}^{a}\frac{1}{u}du\]\[\lim_{a \rightarrow 4^-} \frac{1}{2}\ln|x^2-16|~{\LARGE|}^{a}_{1}\]

Answer 3

oh, all function is improper?

Answer 4

because the function is undefined at x=1 x=2 and on all interval

Answer 5

So there is no area under the curve... so it diverges or what do I put?

Answer 6

I mean the antiderivative is undefined over [1,4)

Answer 7

looks good

Answer 8

as I plug my numbers though, all my values in the log are negative

Answer 9

oh, the absolute value

Answer 10

Correct, but double check:
\[\lim_{x \to 4^-} \ln|x^2-16| \]

Answer 11

something wrong with this?

Answer 12

it should be approaching 4 from the left so that part is fine.

Answer 13

Nope, but the reasoning may be a bit off. The anti-derivatiive is defined on that interval. I just mean to point out it does diverge, but due to the limit I showed above.

Answer 14

why do I suck so much?
\[\lim_{a \rightarrow 4^-} \frac{1}{2}\ln|a^2-16|+\lim_{a \rightarrow 4^-} \frac{1}{2}\ln|1^2-16|\]\[\frac{1}{2}\ln(15)+\lim_{a \rightarrow 4^-} \frac{1}{2}\ln|a^2-16|\]\[\frac{1}{2}\ln(15)+\frac{1}{2}\lim_{a \rightarrow 4^-} \ln|a^2-16|\]\[\frac{1}{2}\ln(15)+\frac{1}{2}\ln(\infty^{-1})\]
yes, it is negative infinity, as I was about to show

Answer 15

"why do I suck so much" is what I didn't remove after I forgot about the absolute value.

Answer 16

Nah, its all good. You just jumped the gun on a claim. It happens a lot to everyone.

Answer 17

yes, my brain is a little off.

Answer 18

ty