Respuesta :
The expression [tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx[/tex] is an illustration of definite integrals
The approximated value of the definite integral is 0.0059
How to evaluate the definite integral?
The definite integral is given as:
[tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx[/tex]
For arctan(x), we have the following series equation:
[tex]\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}[/tex]
Multiply both sides of the equation by x^3.
So, we have:
[tex]x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}} * x^3[/tex]
Apply the law of indices
[tex]x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}[/tex]
[tex]x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}[/tex]
Evaluate the product
[tex]x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}[/tex]
Introduce the integral sign to the equation
[tex]\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =\int\limits^{1/2}_{0} \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}[/tex]
Integrate the right hand side
[tex]\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}} ]\limits^{1/2}_{0}[/tex]
Expand the equation by substituting 1/2 and 0 for x
[tex]\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - [ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{0^{2n + 4}}{2n + 1}} ][/tex]
Evaluate the power
[tex]\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - 0[/tex]
[tex]\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}}[/tex]
The nth term of the series is then represented as:
[tex]T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}[/tex]
Solve the series by setting n = 0, 1, 2, 3 ..........
[tex]T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625[/tex]
[tex]T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238[/tex]
[tex]T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277[/tex]
[tex]T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131[/tex]
..............
At n = 2, we can see that the value of the series has 4 zeros before the first non-zero digit
This means that we add the terms before n = 2
This means that the value of [tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx[/tex] to 4 decimal points is
[tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238[/tex]
Evaluate the difference
[tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762[/tex]
Approximate to four decimal places
[tex]\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059[/tex]
Hence, the approximated value of the definite integral is 0.0059
Read more about definite integrals at:
https://brainly.com/question/15127807
