# Integration and Carbon-14 help

• Jan 24th 2010, 03:52 PM
Integration and Carbon-14 help
Questions:
Integrate e^(x)sec(e^(x))dx

Carbon 14 question: How old is a painting when it has 99.5% of it's original carbon 14? (assuming 5700 is the half-life of carbon 14)

I think you use u-substitution for the 1st one, but I'm not so sure what to do with that, so walk me through it. Also the answer for that is ln lsec(e^(x)+tan(e^(x))l+c (read as logarithm of the absolute value...etc)

For the other one, am I supposed to memorize the constant for carbon-14? Thank you
• Jan 24th 2010, 04:06 PM
General
Quote:

Questions:
Integrate e^(x)sec(e^(x))dx

Substitute $u=e^x$ ---> $du=e^xdx$
$I=\int sec(u) du$

What is the value of the last integral?
• Jan 24th 2010, 04:07 PM
VonNemo19
Quote:

Questions:
Integrate e^(x)sec(e^(x))dx

Carbon 14 question: How old is a painting when it has 99.5% of it's original carbon 14? (assuming 5700 is the half-life of carbon 14)

I think you use u-substitution for the 1st one, but I'm not so sure what to do with that, so walk me through it. Also the answer for that is ln lsec(e^(x)+tan(e^(x))l+c (read as logarithm of the absolute value...etc)

For the other one, am I supposed to memorize the constant for carbon-14? Thank you

For the first. Let $u=e^x$ and recall that $\int\sec{u}du=\ln|\sec{u}+\tan{u}|+C$.
• Jan 24th 2010, 04:08 PM
skeeter
Quote:

Questions:
Integrate e^(x)sec(e^(x))dx

Carbon 14 question: How old is a painting when it has 99.5% of it's original carbon 14? (assuming 5700 is the half-life of carbon 14)

I think you use u-substitution for the 1st one, but I'm not so sure what to do with that, so walk me through it. Also the answer for that is ln lsec(e^(x)+tan(e^(x))l+c (read as logarithm of the absolute value...etc)

For the other one, am I supposed to memorize the constant for carbon-14? Thank you

$\int e^x \sec(e^x) \, dx$

$u = e^x$ ... proceed using substitution.

$y = y_0 e^{kt}$

$\frac{1}{2} = e^{k \cdot 5700}$

solve for $k$, then find the value of t when $y = .995$ , $y_0$ will equal 1, of course.