The expression
is an illustration of definite integrals
The approximated value of the definite integral is 0.0059
<h3>How to evaluate the definite integral?</h3>
The definite integral is given as:
![\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_0%20%7Bx%5E3%20%5Carctan%28x%29%7D%20%5C%2C%20dx)
For arctan(x), we have the following series equation:
![\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}](https://tex.z-dn.net/?f=%5Carctan%28x%29%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%201%7D%7D%7B2n%20%2B%201%7D%7D)
Multiply both sides of the equation by x^3.
So, we have:
![x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}} * x^3](https://tex.z-dn.net/?f=x%5E3%20%2A%20%5Carctan%28x%29%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%201%7D%7D%7B2n%20%2B%201%7D%7D%20%20%2A%20x%5E3)
Apply the law of indices
![x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}](https://tex.z-dn.net/?f=x%5E3%20%2A%20%5Carctan%28x%29%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%201%20%2B%203%7D%7D%7B2n%20%2B%201%7D%7D)
![x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}](https://tex.z-dn.net/?f=x%5E3%20%2A%20%5Carctan%28x%29%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D)
Evaluate the product
![x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}](https://tex.z-dn.net/?f=x%5E3%20%5Carctan%28x%29%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D)
Introduce the integral sign to the equation
![\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}}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%20x%5E3%20%5Carctan%28x%29%5C%20dx%20%3D%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D)
Integrate the right hand side
![\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}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%20x%5E3%20%5Carctan%28x%29%5C%20dx%20%3D%5B%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7Bx%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D%20%5D%5Climits%5E%7B1%2F2%7D_%7B0%7D)
Expand the equation by substituting 1/2 and 0 for x
![\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}} ]](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%20x%5E3%20%5Carctan%28x%29%5C%20dx%20%3D%5B%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7B%281%2F2%29%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D%20%5D%20-%20%5B%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7B0%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D%20%5D)
Evaluate the power
![\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](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%20x%5E3%20%5Carctan%28x%29%5C%20dx%20%3D%5B%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7B%281%2F2%29%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D%20%5D%20-%200)
![\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}}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_%7B0%7D%20%20x%5E3%20%5Carctan%28x%29%5C%20dx%20%3D%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%20%3D%200%7D%20%7B%28-1%29%5En%20%5Ccdot%20%5Cfrac%7B%281%2F2%29%5E%7B2n%20%2B%204%7D%7D%7B2n%20%2B%201%7D%7D)
The nth term of the series is then represented as:
![T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}](https://tex.z-dn.net/?f=T_n%20%3D%20%5Cfrac%7B%28-1%29%5En%7D%7B2%5E%7B2n%20%2B%205%7D%20%2A%20%282n%20%2B%204%29%282n%20%2B%201%29%7D)
Solve the series by setting n = 0, 1, 2, 3 ..........
![T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625](https://tex.z-dn.net/?f=T_0%20%3D%20%5Cfrac%7B%28-1%29%5E0%7D%7B2%5E%7B2%280%29%20%2B%205%7D%20%2A%20%282%280%29%20%2B%204%29%282%280%29%20%2B%201%29%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5E5%20%2A%204%20%2A%201%7D%20%3D%200.00625)
![T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238](https://tex.z-dn.net/?f=T_1%20%3D%20%5Cfrac%7B%28-1%29%5E1%7D%7B2%5E%7B2%281%29%20%2B%205%7D%20%2A%20%282%281%29%20%2B%204%29%282%281%29%20%2B%201%29%7D%20%3D%20%5Cfrac%7B-1%7D%7B2%5E7%20%2A%206%20%2A%203%7D%20%3D%20-0.0003720238)
![T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277](https://tex.z-dn.net/?f=T_2%20%3D%20%5Cfrac%7B%28-1%29%5E2%7D%7B2%5E%7B2%282%29%20%2B%205%7D%20%2A%20%282%282%29%20%2B%204%29%282%282%29%20%2B%201%29%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5E9%20%2A%208%20%2A%205%7D%20%3D%200.00004340277)
![T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131](https://tex.z-dn.net/?f=T_3%20%3D%20%5Cfrac%7B%28-1%29%5E3%7D%7B2%5E%7B2%283%29%20%2B%205%7D%20%2A%20%282%283%29%20%2B%204%29%282%283%29%20%2B%201%29%7D%20%3D%20%5Cfrac%7B-1%7D%7B2%5E%7B11%7D%20%2A%2010%20%2A%207%7D%20%3D%20-0.00000634131)
..............
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
to 4 decimal points is
![\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_0%20%7Bx%5E3%20%5Carctan%28x%29%7D%20%5C%2C%20dx%20%3D%200.00625%20-%200.0003720238)
Evaluate the difference
![\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_0%20%7Bx%5E3%20%5Carctan%28x%29%7D%20%5C%2C%20dx%20%3D%200.0058779762)
Approximate to four decimal places
![\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B1%2F2%7D_0%20%7Bx%5E3%20%5Carctan%28x%29%7D%20%5C%2C%20dx%20%3D%200.0059)
Hence, the approximated value of the definite integral is 0.0059
Read more about definite integrals at:
brainly.com/question/15127807