Question

@Michele_Laino

Answer 1

why?

Answer 2

plese note that the solution of your integral is:
\[y=F(x)=18[\arcsin \left( \frac{ x }{ 6 } \right)+\frac{ x }{ 6 }\sqrt{1-\frac{ x ^{2} }{ 36 }}]+9 \pi\]
so?

Answer 3

no, you have to check the existence of radical, and the existence of the arcsin function, so?

Answer 4

for example, radical exists if and only if \[1-\frac{ x ^{2} }{ 36 }\ge 0\]
and arcsin exists if and only if:
\[-1\le \frac{ x }{ 6 }\le +1\]

Answer 5

please solve the second double inequality

Answer 6

wait, please!

Answer 7

please insert x=-6 into my function, and write what you get

Answer 8

sorry, please compute F(-6)

Answer 9

no, F(-6)=0

Answer 10

now, please compute F(6)

Answer 11

since, the domain of F(x) is the set:
\[-6 \le x \le6\]

Answer 12

what is F(6)? please?

Answer 13

6

Answer 14

no, sorry
\[F(6)=18 \pi\]
so the range of F is?

Answer 15

wait so i was right?

Answer 16

what was your option?

Answer 17

B

Answer 18

ok!

Answer 19

wait so i was right??

Answer 20

Yes! please, keep in mind we have to justify our answer everytime!