Mathematics, 16.08.2020 14:01, nisha87
Use a power series to approximate the definite integral, I, to six decimal places. 0.1 x arctan(5x) dx 0.
Answers: 3
![\int^{0.1}_{0} \ x \ arctan (5x) \ dx = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ 5^{2n +1}}{2n+1} \ [(0.1)^{2n+3}]](/tpl/images/0723/1184/9de97.png)
![\int^{0.1}_{0} \ x \ arctan (5x) \ dx = [\dfrac{5 \times (0.1)^3}{1.3}-\dfrac{5^3(0.1)^3}{3.5}+\dfrac{5^5(0.1)^7}{5.7}-\dfrac{5^7(0.1)^9}{7.9}+ ...]](/tpl/images/0723/1184/a2e87.png)
![\int^{0.1}_{0} \ x \ arctan (5x) \ dx = [\dfrac{1}{600}-\dfrac{1}{1200}+\dfrac{1}{112000}-\dfrac{1}{806400}+ ...]](/tpl/images/0723/1184/ea8f7.png)
to six decimal places
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Use a power series to approximate the definite integral, I, to six decimal places. 0.1 x arctan(5x)...
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