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3 years ago

How many gifts were given in the 12 days of christmas brainly.

1 answer:

3 0

The 12 days of Christmas is a very popular song. For us to know how many gifts were given in the 12 days of Christmas, we have to add up all the gifts;

She was given 1 partridge in a pear tree for each of the 12 days: So we multiply 1 x 12=12.

She was given 2 turtle doves on the last 11 days. So we shall multiply 2 x 11 = 22.

She was given 3 French hens on the last 10 days. So we shall multiply 3 x 10 = 30.

She was given 4 calling birds on the last 9 days. So we shall multiply 4 x 9 = 36.

She was given 5 gold rings on the last 8 days. So we shall multiply 5 x 8 = 40.

She was given 6 geese a-laying on the last 7 days. So we shall multiply 6 x 7 = 42.

She was given 7 swans a-swimming on the last 6 days. So we shall mutliply7 x 6 = 42.

She was given 8 maids a-milking on the last 5 days. So we shall multiply 8 x 5 = 40.

She was given 9 ladies dancing on the last 4 days. So we shall multiply9 x 4

= 36.

She was given 10 lords a-leaping on the last 3 days of Christmas. so we multiply 10 x 3=30.

She was given 11 pipers piping on the last 2 days. So we shall multiply 11 x 2=22.

She was given 12 drummers drumming on the last day. So we shall multiply 12 x 1=12

Therefore you sum up all the totals

12+30+22+12....... = 364gifts

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Use series to approximate the definite integral i to within the indicated accuracy. i = 1/2 x3 arctan(x) dx 0 (four decimal plac

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The expression $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ 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:

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$

For arctan(x), we have the following series equation:

$\arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}}$

Multiply both sides of the equation by x^3.

So, we have:

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1}}{2n + 1}} * x^3$

Apply the law of indices

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 1 + 3}}{2n + 1}}$

$x^3 * \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Evaluate the product

$x^3 \arctan(x) = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Introduce the integral sign to the equation

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =\int\limits^{1/2}_{0} \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}}$

Integrate the right hand side

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{x^{2n + 4}}{2n + 1}} ]\limits^{1/2}_{0}$

Expand the equation by substituting 1/2 and 0 for x

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - [ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{0^{2n + 4}}{2n + 1}} ]$

Evaluate the power

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx =[ \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}} ] - 0$

$\int\limits^{1/2}_{0} x^3 \arctan(x)\ dx = \sum\limits^{\infty}_{n = 0} {(-1)^n \cdot \frac{(1/2)^{2n + 4}}{2n + 1}}$

The nth term of the series is then represented as:

$T_n = \frac{(-1)^n}{2^{2n + 5} * (2n + 4)(2n + 1)}$

Solve the series by setting n = 0, 1, 2, 3 ..........

$T_0 = \frac{(-1)^0}{2^{2(0) + 5} * (2(0) + 4)(2(0) + 1)} = \frac{1}{2^5 * 4 * 1} = 0.00625$

$T_1 = \frac{(-1)^1}{2^{2(1) + 5} * (2(1) + 4)(2(1) + 1)} = \frac{-1}{2^7 * 6 * 3} = -0.0003720238$

$T_2 = \frac{(-1)^2}{2^{2(2) + 5} * (2(2) + 4)(2(2) + 1)} = \frac{1}{2^9 * 8 * 5} = 0.00004340277$

$T_3 = \frac{(-1)^3}{2^{2(3) + 5} * (2(3) + 4)(2(3) + 1)} = \frac{-1}{2^{11} * 10 * 7} = -0.00000634131$

..............

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 $\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx$ to 4 decimal points is

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.00625 - 0.0003720238$

Evaluate the difference

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0058779762$

Approximate to four decimal places

$\int\limits^{1/2}_0 {x^3 \arctan(x)} \, dx = 0.0059$

Hence, the approximated value of the definite integral is 0.0059

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2 years ago