Answer:
definite integral I = 0.001591 to six decimal places
Step-by-step explanation:
The definite integral is given as:
[tex]\int ^{0.1}_{0} \ x \ arctan (5x) \ dx[/tex]
For arctanx , the power series is in the order [tex]x - \dfrac{x^3}{3}+ \dfrac{x^5}{5}-\dfrac{x^7}{7}+...[/tex]
[tex]arc tan \ x = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ x ^{2n+1}}{2n +1}[/tex]
The next step is to substitute the value of 5x for x in the above equation;
So,
[tex]arc tan \ (5x) = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ (5x) ^{2n+1}}{2n +1}[/tex]
To multiply both sides by (x); we have
[tex]x\ arc tan \ (5x) = x \ \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ 5 ^{2n+1} \times x^{2n+1}}{2n +1}[/tex]
[tex]x\ arc tan \ (5x) = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ 5 ^{2n+1} \times x^{2n+2}}{2n +1}[/tex]
Taking the integral on both sides with respect to x;
[tex]\int^{0.1}_{0} \ x \ arctan (5x) \ dx = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ 5^{2n +1}}{2n+1} \ \int ^{0.1}_0 x^{2n+2} \ dx[/tex]
[tex]\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}][/tex]
[tex]\int^{0.1}_{0} \ x \ arctan (5x) \ dx = \sum \limits ^{\infty}_{n=0} \dfrac{(-1)^n \ 5^{2n +1} \times (0.1)^{2n+3} }{(2n+1)(2n+3)}[/tex]
[tex]\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}+ ...][/tex]
[tex]\int^{0.1}_{0} \ x \ arctan (5x) \ dx = [\dfrac{1}{600}-\dfrac{1}{1200}+\dfrac{1}{112000}-\dfrac{1}{806400}+ ...][/tex]
[tex]\int^{0.1}_{0} \ x \ arctan (5x) \ dx =1.591 \times 10^{-3}[/tex]
[tex]\mathbf{\int^{0.1}_{0} \ x \ arctan (5x) \ dx =0.001591}[/tex] to six decimal places
