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**Prove It** OK, I've done it, but I don't think you'll like the solution...

For this to work, we need to make use of the Mean Value Theorem.

i.e. for some function $\displaystyle F(x)$, which is continuous on $\displaystyle x \in [a, b]$, we have

$\displaystyle f(c) = \frac{F(b) - F(a)}{b - a}$, where $\displaystyle a < c < b$ and $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

We can rearrange the formula so that it is of the form

$\displaystyle (b - a)f(c) = F(b) - F(a)$.

For this function, $\displaystyle f(x) = \ln{x}$, suppose we take some rectangular strip, between $\displaystyle x_{k - 1}$ and $\displaystyle x_k$.

Then this strip's area is

$\displaystyle (x_k - x_{k - 1})f(c)$, for some $\displaystyle c \in (x_{k - 1}, x_k)$.

Using the Mean Value Theorem, we can rewrite this as

$\displaystyle F(x_k) - F(x_{k - 1})$, where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$.

Clearly, since $\displaystyle f(x) = \ln{x}$, this means $\displaystyle F(x) = x\ln{x} - x$.

So the area of one strip is...

$\displaystyle x_k\ln{x_k} - x_k - (x_{k - 1}\ln{x_{k - 1}} - x_{k - 1})$

$\displaystyle = x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1}$.

If we have $\displaystyle n$ strips, then the Area between $\displaystyle x_0$ and $\displaystyle x_n$ is

$\displaystyle A = \sum_{k = 1}^n{(x_k\ln{x_k} - x_k - x_{k - 1}\ln{x_{k - 1}} + x_{k - 1})}$

$\displaystyle = x_n\ln{x_n} - x_n - x_0\ln{x_0} + x_0$, since this is a telescopic series.

Since $\displaystyle x_0 = 1$ and $\displaystyle x_n = t$, you have

$\displaystyle A = t\ln{t} - t - 1\ln{1} + 1$

$\displaystyle = t\ln{t} - t + 1$.

To fully complete this, you would need to note that you make $\displaystyle n \to \infty$ and then take the limit.

Can you evaluate

$\displaystyle \lim_{n \to \infty}(t\ln{t} - t + 1)$?