 # Express the definite integral as an infinite series and find its value to within Globokim8 2021-11-05 Answered
Express the definite integral as an infinite series and find its value to within an error of at most
${10}^{-4}$
${\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx$
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Term-by-Term Differentiation and Integration
$F\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{\left(x-c\right)}^{n}$
has radius of convergence $R>0$. Then F is differentiable on $\left(c-R,c+R\right)$. Furtehermore, we can integrate and differentiate term by term. For $x\in \left(c-R,c+R\right)$
${F}^{\prime }\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}n{a}_{n}{\left(x-c\right)}^{n-1}$
$\int F\left(x\right)dx=A+\sum _{n=0}^{\mathrm{\infty }}\frac{{a}_{n}}{n+1}{\left(x-c\right)}^{n+1}$ (A any constant)
These series have the same radius of convergence R
Here we need to find the value of $F\left(x\right)={\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx$ we will use the expansion of the $\mathrm{cos}x$
From table 2, We have Maclaurin Series $f\left(x\right)=\mathrm{cos}x=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots$ converges for for all x here x is replaced with ${x}^{2}$
${\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{\left({x}^{2}\right)}^{2n}}{\left(2n\right)!}$
${\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{4n}}{\left(2n\right)!}dx$
${\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\in }ft{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}$
Now we need to find out F(1) error less than 0.0001
${\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}$
Above is alternating series with