e^[int (0.5(1 + t)^(-2))] between 0 and 9

[eliseo9]

Let me try to explain the problem:

I need to integrate 0.5(1 + t)^(-2) with respect to t, between the values of 0 and 9.

The answer will be the exponent of e. I know what the answer will be "1.5683", but I don't know how the answer is derived.

 

[tkhunny]

\(\displaystyle \L\,\int{\frac{1}{(1+t)^{2}}}\,dt\)

You can't do that?

It's just a power rule.

The fact that you are even hinting at variable interest rate accumulations suggests to me that you have seen more than just a tiny chunk of calculus. Are you SURE you can't do it?

 

[eliseo9]

it's been years since i have done this in college, and i haven't been putting it into practice until now. its more rust than anything.

 

[galactus]

Hello elise:

Is this your integral:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{9}\frac{1}{(1+t)^{2}}dt\)

Let \(\displaystyle \L\\u=1+t,\;\ then \;\ du=dt\)

Change variables of integration from 1 to 10 because letting t=1 and 9, we have u=1+9=10 and u=1+0=1

\(\displaystyle \L\\\frac{1}{2}\int_{1}^{10}\frac{1}{u^{2}}du\)

Integrate:

\(\displaystyle \L\\\frac{1}{2}\left[\frac{-1}{u}\right]\)


\(\displaystyle \L\\\frac{1}{2}(\frac{-1}{10}-\frac{-1}{1})=\frac{9}{20}=0.45\)

So, you have \(\displaystyle \H\\e^{\frac{9}{20}}\approx{1.5683}\)

 

[tkhunny]

it's been years since i have done this in college, and i haven't been putting it into practice until now. its more rust than anything.

OK, just checking. It is a variable interest problem, isn't it?

 

[eliseo9]

thank you and yes