Can someone check if my answer is correct because I'm not sure if I have done it right? I think I have, but I'm not sure :)
Thanks in advance
Consider the function $f(x)=2x+2\ln(1+x)$ for $x>-1$ and centered at $0$.
Show that $$|f(x)-T_3(x)|\leq\frac{x^4}{2}$$ for all $x\geq0$. Where $T_3(x)=4x-x^2+\frac{2}{3}x^3$
Use it to show the following $$ |\int_0^a{2x+2\ln(1+x)}dx-2a^2+\frac{a^3}{3}-\frac{a^4}{6}|\leq \frac{1}{10}a^5 $$
Answer:
To show the first inequality $$|f(x)-T_3(x)|\leq\frac{x^4}{2}$$ I use the The Remainder Estimation Theorem: $$|f(x)-T_n(x)| \leq \frac{M_n}{(n+1)!}x^{n+1}$$ Where $n=3$ $$|f^{(3+1)}(x)| \leq M_3$$ $$|f^{(4)}(x)|=\frac{12}{(x+1)^4}$$ And for all $x \geq 0$, we have that $$M_3=12$$ So we have $$|f(x)-T_3(x)| \leq \frac{12}{4!}x^{4}=\frac{12}{24}x^{4}=\frac{1}{2}x^{4}$$
And to show the last inequality where $$\int_0^a{T_3(x)}dx=2a^2-\frac{a^3}{3}+\frac{a^4}{6}$$ is an approximation of the integral $$\int_0^a{2x+2\ln(1+x)}dx$$ so we have that $$ |\int_0^a{2x+2\ln(1+x)}dx-2a^2+\frac{a^3}{3}-\frac{a^4}{6}|\leq \int_0^a{\frac{x^4}{2}}dx=[\frac{x^5}{10}]_0^a=\frac{a^5}{10}-\frac{0^5}{10}=\frac{a^5}{10} $$ We have now shown the last inequality