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$I=\int_0^{1.6}\frac {1}{1+x^4} dx$ by using generalized trapozoidal rule $n=8$ the final answer don't equal the correct answer .I need the final answer and how can i solve it ?

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mahmoud afefey
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2 Answers2

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$$\int_0^{1.6}\frac{1}{x^4+1}dx\approx (1.6-0)*\frac{\frac{1}{1.6^4+1}+\frac{1}{0^4+1}}{2}\\\approx 1.6*\frac{\frac{1}{1.6^4+1}+1}{2}\\\approx1.6*\frac{2+1.6^4}{2*(1.6^4+1)}\\\approx 0.90590976488 $$ NOTE:This is an approximation,the real answer is $1.0342$ but the trapezoid rule gives a number close to the result

kingW3
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If you do it in two steps $$I=\int_0^{1.6}\frac{1}{x^4+1}dx=\int_0^{0.8}\frac{1}{x^4+1}dx+\int_{0.8}^{1.6}\frac{1}{x^4+1}dx$$ and apply, for each interval, exactly what kingW3 wrote, you would arrive to $1.02049$ for the exact value of $1.0342$. This shows you, I hope, the impact of the number of trapezes.

Claude Leibovici
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