Question

$\huge{ \boxed{ \bold \red{ \int(6 \sqrt{x} - 1)dx}}}$




❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

Answer:

a^2 + a + 1 = 0, a^2 + 1/a^2 is a

a) positive integer

b) positive fraction

c) negative integer

d) negative fraction

Answer 2

$\bold \red{\int{\left(6 \sqrt{x} - 1\right)d x}}$

Integrate term by term:

${\bold \red{\color{red}{\int{\left(6 \sqrt{x} - 1\right)d x}} = \color{red}{\left(- \int{1 d x} + \int{6 \sqrt{x} d x}\right)}}}$

Apply the constant rule

$\bold \red{\int c\, dx = c x \: with \: c=1:}$

${\bold \red{\int{6 \sqrt{x} d x} - \color{red}{\int{1 d x}} = \int{6 \sqrt{x} d x} - \color{red}{x}}}$

Apply the constant multiple rule

${\bold \red{\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=6 and f{\left(x \right)} = \sqrt{x}:}}$

${ \bold \red{- x + \color{red}{\int{6 \sqrt{x} d x}} = - x + \color{red}{\left(6 \int{\sqrt{x} d x}\right)}}}$

Apply the power rule

${ \bold \red{ \int x^{n}\, dx = \frac{x^{n + 1}}{n + 1} \left(n \neq -1 \right) \: with \: n=\frac{1}{2}:}}$

${ \bold \red{- x + 6 \color{red}{\int{\sqrt{x} d x}}=- x + 6 \color{red}{\int{x^{\frac{1}{2}} d x}}=- x + 6 \color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}=- x + 6 \color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}}$

Therefore,

$\bold \red{\int{\left(6 \sqrt{x} - 1\right)d x} = 4 x^{\frac{3}{2}} - x}$

Add the constant of integration:

$\bold \red{\int{\left(6 \sqrt{x} - 1\right)d x} = 4 x^{\frac{3}{2}} - x+C}$