Let $I(a)=\int_0^ax^2\sin(x^3)dx$. Express $I(a)$ as (power) series and calculate $I(0.7)$ using the first two terms. Calculate error of $I(0.7)$.
It is known that $$\sin x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}.$$ Thus $$x^2\sin(x^3)=\sum_{k=0}^\infty(-1)^k\frac{x^{6k+5}}{(2k+1)!}$$ and $$\int_0^a x^2\sin(x^3)dx=\int_0^a \sum_{k=0}^\infty(-1)^k\frac{x^{6k+5}}{(2k+1)!}dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^a x^{6k+5}dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{a^{6k+6}}{6k+6}$$
$$I(0.7)=\frac{0.7^6}{6}-\frac{0.7^{12}}{12\cdot3!}+\dots$$
How would one calculate the error estimate? How would one conclude that the series expansion converges?
You can take $x^3 = t$ and then integrate this expression exactly without using power series. And then you can calculate %error as the %deviation form exact answer.