Question

Help me please!!the question at below

Answer 1

cannot write the question bcoz its to long..T.T

Answer 2

what have u tried ?
tried splitting the integral ?

Answer 3

i have try solv the question and i get -17/3 for the answer..@hartnn is it the answer is correct?

Answer 4

*solve

Answer 5

h(x) part is easy
-8

Answer 6

f(x) part

-1/3 (-1/2 + 5) =..

Answer 7

sorry -5

Answer 8

11/6 - 8 = (11-48)/8 = - 37/8

Answer 9

how you got -17/3?

Answer 10

h(x) part is =-8 and f(x) part is=7/3

Answer 11

let me calculate f part again

Answer 12

6 to 9 of (f(x)) = 1/2 +5

so 9 to 6 of f(x) = -1/2 -5
ok?

Answer 13

Omg, I LOVE how colorful the question is!!!

Answer 14

ikr!

Answer 15

-1/3 (-11/2) = -11/6

thats what i got for f(x)

Answer 16

typo : +11/6 **

Answer 17

you're getting how i am calculating f part ?

Answer 18

ya i got it..the orginal function is 2 multiply f(x),so to inverse it,we have to divide f(x) by 2..right?

Answer 19

you mean from the limits of 6 to 7, ?
then yes :)

Answer 20

thanks for fixing my mistake

Answer 21

if i have f(x)+2,is it the inverse is f(x)-2?

Answer 22

you want inverse of [f(x)+2] ??

and no its not f(x)-2

Answer 23

f(x)-2 seems like the conjugate of f(x)+2

Answer 24

if you're referring something like this :
\(\int \limits_a^b [f(x)+2] dx= k\)
then
\(\int \limits_a^b [f(x)] dx\ne k-2\)

Answer 25

\(\int \limits_a^b [f(x)] dx = k-2(b-a)\)

Answer 26

\[\int\limits_{b}^{a} [f(x)+2] dx= -k\]

Answer 27

\(\int\limits_{b}^{a} [f(x)+2] dx= -k \\ \int\limits_{b}^{a} [f(x)] dx + 2[x]_b^a= -k \\ \int\limits_{b}^{a} [f(x)] dx +2(a-b)= -k \\ \int\limits_{b}^{a} [f(x)] dx= -k-2(a-b)\)

Answer 28

ask if any doubt in any step...

Answer 29

2 is a constant right?

Answer 30

yes :)

Answer 31

we're just "adding" on x's to all values inside the integrand.

Answer 32

Therefore because an x was "added" onto the constant 2, we were able to evaluate it at a and b