Show that the error is less than $C_0a^4/4$

Question

Consider the function $f(x)=\ln(2+\sin(x))$, for $x\in \mathbb{R}$. Where $$ T_2(x)=\ln(2)+\frac{1}{2}x-\frac{1}{8}x^2 $$ is the Taylor polynomial of degree 2 centered at $x_0=0$. Find a constant $C_0>0$ such that $$|f(x)-T_2(x)|\leq C_0x^3$$ for all $x\in[0,1]$

Then I use the remainder estimation theorem $$|R_nf(x)|\leq \frac{M_n}{(n+1)!}|x-x_0|^{n+1}$$ where $$\max{\{|f^{(n+1)}(t)| : t\in [x_0,x]\}}\leq M_n$$

I find that $$|f^{(3)}(t)|=\frac{-2 \cdot ((\cos(t))^2+(\sin(t)-1) \cdot (\sin(t)+2))\cdot|\cos(t)|}{(\sin(t)+2)^3}$$

Then I find out that the maximum value for $|f^{(3)}(t)|$ is $\frac{1}{4}$ which can be obtained at t=0. so we have that $$|f^{(3)}(t)|\leq M_2 =\frac{1}{4}$$ Then we have $$|f(x)-T_2(x)|\leq \frac{\frac{1}{4}}{3!}x^3=\frac{\frac{1}{4}}{6}x^3=\frac{1}{24}x^3$$ So that means $C_0=\frac{1}{24}$.

Then I have to use it so I can find a approximation of the integral $$I(a)=\int_0^a \ln(2+\sin(x))dx$$ for $0\leq a \leq 1$ as a polynomial in $a$ and explain that the error is less than $C_0a^4/4$

To find the approximation of the integral, I find the integral of the Taylor polynomial: $$ T_2(x)=\ln(2)+\frac{1}{2}x-\frac{1}{8}x^2 $$ so the approximation of the integral is the following: $$ \int_0^a \ln(2)+\frac{1}{2}x-\frac{1}{8}x^2dx=0.693a+0.25a^2-0.041a^3 $$ Now I have to show that the error is less than $C_0a^4/4$, but I don't know how to do it.

Thanks in Advance.

Answer 1

$$\begin{align}\int_0^a\ln(2+\sin x)dx&=\int_0^a[T_2(x)+(\ln(2+\sin x)-T_2(x))]dx\\&=\int_0^aT_2(x)dx+\int_0^a(\ln(2+\sin x)-T_2(x))dx\\&\le\int_0^aT_2(x)dx+\int_0^aC_0x^3dx\\&=\int_0^aT_2(x)dx+\frac{C_0}4a^4\end{align}$$

Answer 2

Just note that, according to your own calculations,

$$ \int_0^a |f(x)-T_2(x)| dx \leq \int_0^a C_0 x^3 dx = \frac{C_0 a^4}{4}, $$

where $C_0$ is the maximum value of $f'''$ over the interval $[0,a]$.