Question

How would you attack this? Calculus

Answer 1

is this 1-x/ sqrtx ?

Answer 2

yes

Answer 3

if so, split into 1/sqrtx - sqrtx

Answer 4

why is sqrt x by itself?

Answer 5

then power rule to each term

Answer 6

x/sqrtx = sqrtx

Answer 7

1-1/2 = 1/2

Answer 8

\[\int\limits_{1}^{4} \frac{ 1-x }{ x }\]

Answer 9

not a square rot on the bottom/

Answer 10

I still do not understand why the sqrt of x is by itself. Sorry

Answer 11

\[\int\limits_{1}^{4} \frac{1-x}{x}~dx ~~\Longrightarrow~~\int\limits_{1}^{4} \frac{1}{x}-\frac{x}{x}~dx \]

Answer 12

yes

Answer 13

You can draw it will be easier for you

Answer 14

didnt you have a square root?

Answer 15

yes

Answer 16

sorry

Answer 17

\[\int\limits_{1}^{4} \frac{1}{x}-1~dx \]then apply the power rule to the 2nd term, and the other rule,
\[\int\limits_{ }^{ }~1/x~~~dx=~\ln \left| x \right|+C\]

Answer 18

\[\int\limits_{1}^{4} \frac{ 1-x }{ sqrt{x} }\]

Answer 19

So sorry caused confusion

Answer 20

lol

Answer 21

\[\int\limits_{1}^{4} \frac{1-x}{\sqrt{x}}~dx \]\[\int\limits_{1}^{4} \frac{1}{\sqrt{x}}- \frac{x}{\sqrt{x}}~dx \]\[\int\limits_{1}^{4}x^{-1/2}-x^{1/2}~dx \]

Answer 22

then power rule.

Answer 23

I get it, I LOVE YOU GUYS ONCE AGAIN

Answer 24

okay, yw

Answer 25

Distribute the negative half into the parenthesis