Taylor Remainder Term

On page 756 (of Stewart's Single Variable Calculus Early Transcendentals), the margin has the following information:
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.

Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.

Note that Taylor's Inequality follows from Lagrange's form of the remainder term.

Let's do a similar problem to #17 on this the 2010 final exam.

$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder