Question

Please solve this ASAP ​

Answer 1

Question:

$\sf\displaystyle \int \: \dfrac{ {x}^{3} }{ {x}^{2} + 1}$

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Adding and subtracting x

$\sf \large= > \displaystyle \int \: \dfrac{ {x}^{3} - x + x }{ {x}^{2} + 1 } dx$

$\sf\large= > \displaystyle \int \: \dfrac{x( {x}^{2} + 1) - x}{ {x}^{2} + 1 } dx$

$\sf\large= >\displaystyle \int \bigg(\dfrac{x( {x}^{2} + 1)}{ {x}^{2} + 1 } - \dfrac{x}{ {x}^{2} + 1} \bigg)dx$

$\sf\large= >\displaystyle \int\bigg(x - \dfrac{x}{ {x}^{2} + 1}\bigg)dx$

$\sf\large= > \displaystyle \int \: xdx - \int \dfrac{x}{ {x}^{2} + 1 } dx$

$\sf\boxed{ {x}^{n} dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}$

$\sf\large= \dfrac{ {x}^{2} }{2} - \dfrac{1}{2}log( {x}^{2} + 1) + c$

Answer 2

Step-by-step explanation:

number of people who do not show up (x) is a binomial variable with probability 0.2.

probability 0, 1 or 2 are no shows is:

C(31,0) 0.2^0 0.8^31 + C(31,1)0.2^1 0.8^30 + C(31,2) 0.2^2 0.8^29

where C(31,x) = number of combinations of 31 items taking x at a time.

= 0.00099 + 0.00768 + 0.0288 = 0.037