Using Maclaurin polynomials to approximate definite integrals?

Question

Use the 4th degree Maclaurin polynomial:

$$\cos(x^2)=\sum _{k=0}^{\infty}(−1)^k\frac {(x^{4k})}{(2k)!}$$

to approximate the definite integral:

$$ \int _0^1 \cos(x^2)dx$$

So the 4th degree polynomial = 1-(1/2)x^4 I think? So $$ \int _0^1 \cos(x^2)dx = 1-(1/2)x^4.$$ Now what? Is it $$1-(1/2)(1)^4-1-(1/2)(0)^4=-0.5$$ I know I did something wrong because the integral ~ 0.9045...

Answer 1

You have $$ \int_{0}^{1}\cos\left(x^2\right)\text{d}x=\sum_{k=0}^{+\infty}\int_{0}^{1}\left(-1\right)^k \left(\frac{x^{4k}}{(2k)!}\right)\text{d}x=\sum_{k=0}^{+\infty}\left(-1\right)^k\frac{1}{(2k)!(4k+1)} $$ So

$$ \int_{0}^{1}\cos\left(x^2\right)\text{d}x\approx 1-\frac{1}{10}+\frac{1}{216}=0.9046$$