1
$\begingroup$

Evaluate $\displaystyle \frac{1}{5000}\int_0^{100} t (1.1)^t \mathrm d t$.

What am I doing wrong with tabular integration for this problem?

Im using $t$ for the derivative term (so it ends up as $t\to1\to0)$ and integrating with $(1.1)^t$ so the first term is: $\displaystyle \frac{(1.1)^{t+1}}{t+1}$, and the second is $\displaystyle \frac{(1.1)^{t+2}}{(t+2)(t+1)}$.

Combining terms and solving gives me approximately $3000$ which is wrong.

$\endgroup$
1
  • $\begingroup$ Have you taken care of the $\pm$ signs to the left of the tabular "environment"? $\endgroup$
    – Frenzy Li
    Dec 1 '12 at 2:25
4
$\begingroup$

You are using a false integration formula to integrate $1.1^t$. There is a big difference between a function of the shape $t^k$, where $k$ is a constant, and a function of the shape $a^t$, where $a$ is a constant.

Except for the case $k=-1$, $\dfrac{1}{k+1}t^{k+1}$ is an indefinite integral of $t^k$. However, this formula does not apply when the variable is in the exponent.

We now proceed to find an antiderivative of $1.1^t$. Note that $1.1^t=e^{(\ln(1.1) t}$. Integrate. By using the substitution $u=(\ln(1.1))t$, or otherwise, we get that one indefinite integral is $\dfrac{1}{\ln(1.1)}e^{(\ln(1.1)) t}$. This can be rewritten as $\dfrac{1}{\ln(1.1)}(1.1^t)$.

$\endgroup$
1
  • $\begingroup$ Thank you, I understand now. $\endgroup$ Dec 1 '12 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.