Estimate-0-0-5-1-x-4-dx-with-an-error-0-0001-

Question Number 144186 by cherokeesay last updated on 22/Jun/21

E s t i m a t e \int_{0}^{0.5} \sqrt{1 + x^{4}} d x

w i t h a n e r r o r 0.0001

Answered by Dwaipayan Shikari last updated on 23/Jun/21

\int \sqrt{1 + x^{4}} d x

= \int Σ \frac{{(- \frac{1}{2})}_{n}}{n!} {(- 1)}^{n} x^{4 n} d x

= \frac{x}{4} \underset{n = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{n}}{n!} {(- 1)}^{n} \frac{x^{4 n}}{n + \frac{1}{4}} = x Σ \frac{{(- \frac{1}{2})}_{n}}{n!} \frac{x^{4 n} {(\frac{1}{4})}_{n}}{{(\frac{5}{4})}_{n}} {(- x^{4})}^{n}

= x_{2} F_{1} (- \frac{1}{2}, \frac{1}{4} ∣ \frac{5}{4}; - x^{4}) + C

\int_{0}^{\frac{1}{2}} \sqrt{1 + x^{4}} d x = \frac{1}{2}_{2} F_{1} (- \frac{1}{2}, \frac{1}{4} ∣ \frac{5}{4}; - \frac{1}{16})

Commented by cherokeesay last updated on 23/Jun/21

t h a n k y o u s i r .

Answered by mathmax by abdo last updated on 23/Jun/21

we have {(1 + x)}^{α} = 1 + α x + \frac{α (α - 1)}{2} x^{2} + \dots . \Rightarrow

{(1 + x)}^{\frac{1}{2}} = 1 + \frac{x}{2} - \frac{1}{8} x^{2} + \dots \Rightarrow 1 + \frac{x}{2} - \frac{x^{2}}{8} ⩽ \sqrt{1 + x} ⩽ 1 + \frac{x}{2}

we change x by x^{4} \Rightarrow 1 + \frac{x^{4}}{2} - \frac{x^{8}}{8} ⩽ \sqrt{1 + x^{4}} ⩽ 1 + \frac{x^{4}}{2} \Rightarrow

\int_{0}^{\frac{1}{2}} (1 + \frac{x^{4}}{2} - \frac{x^{8}}{8}) dx ⩽ \int_{0}^{\frac{1}{2}} \sqrt{1 + x^{4}} dx ⩽ \int_{0}^{\frac{1}{2}} (1 + \frac{x^{4}}{2}) dx we have

\int_{0}^{\frac{1}{2}} (1 + \frac{x^{4}}{2} - \frac{x^{8}}{8}) dx = {[x + \frac{1}{10} x^{5} - \frac{1}{72} x^{9}]}_{0}^{\frac{1}{2}} = \frac{1}{2} + \frac{1}{10 . 2^{5}} - \frac{1}{72 . 2^{9}}

\int_{0}^{\frac{1}{2}} (1 + \frac{x^{4}}{2}) x = {[x + \frac{1}{10} x^{5}]}_{0}^{\frac{1}{2}} = \frac{1}{2} + \frac{1}{10 . 2^{5}} \Rightarrow

\frac{1}{2} + \frac{1}{10 . 2^{5}} - \frac{1}{72 . 2^{9}} ⩽ \int_{0}^{\frac{1}{2}} \sqrt{1 + x^{4}} dx ⩽ \frac{1}{2} + \frac{1}{10 . 2^{5}}

the best value of this integral is

v_{0} = \frac{1}{4} + \frac{1}{10 . 2^{6}} - \frac{1}{72 . 2^{10}} + \frac{1}{4} + \frac{1}{10 . 2^{6}}

= \frac{1}{2} + \frac{1}{5 . 2^{6}} - \frac{1}{72 . 2^{10}}