Question

If f(x) is even and
6
(integral of)(f(x) − 4) dx = 8
−6

find
6
(integral of)f(x) dx.
0

Answer 1

Given that \(f(x)\) is even, you know that
\[\int_{-a}^af(x)~dx=2\int_0^a f(x)~dx\]
for \(a\) in the proper domain.

Answer 2

so 16?

Answer 3

its not 16 ;/

Answer 4

you're not accounting for the -4

Answer 5

the transformation changes the answer is just don't know how; i know it makes the function go down by 4 so does that expand the area by 4?

Answer 6

\[\int_{-6}^6(f(x)-4)~dx=2\int_0^6f(x)-4\int_{-6}^6dx\]

Answer 7

is it 12?

Answer 8

my professor ddnt go over how certain transformations affect the area :/

Answer 9

\[\large \int_{-6}^{6} (f(x)-4)dx = \int_{-6}^{6} f(x) dx + \int_{-6}^{6} (-4)dx\]

\[\large \int_{-6}^{6} (f(x)-4)dx = 2*\int_{0}^{6} f(x) dx + 2*\int_{0}^{6} (-4)dx\]

\[\large \int_{-6}^{6} (f(x)-4)dx = ???\]

Answer 10

this only works because f(x) is even

so is the constant function g(x) = -4

Answer 11

hmmmm..?

Answer 12

I guess now that I look back, you don't need to go to \[\Large 2*\int_{0}^{6} (-4)dx\] but oh well

Answer 13

oh wait, I'm thinking in reverse, one sec

Answer 14

okay! no problem!

Answer 15

\[\large \int_{-6}^{6} (f(x)-4)dx = \int_{-6}^{6} f(x) dx + \int_{-6}^{6} (-4)dx\]

\[\large 8 = \int_{-6}^{6} f(x) dx + \int_{-6}^{6} (-4)dx\]

\[\large 8 - \int_{-6}^{6} (-4)dx = \int_{-6}^{6} f(x) dx \]

\[\large \int_{-6}^{6} f(x) dx = 8 - \int_{-6}^{6} (-4)dx\]

\[\large \int_{-6}^{6} f(x) dx = ???\]

Answer 16

umm hmm… its not 12 right?

Answer 17

what is the value of

\[\large \int_{-6}^{6} (-4)dx\]

Answer 18

0

Answer 19

i believe?

Answer 20

no

Answer 21

draw the function y = -4

Answer 22

place vertical lines at x = -6 and x = 6

Answer 23

the integral from -6 to +6 for the function g(x) = -4 is simply the area between x = -6 and x = 6 and also between the x axis and the function curve y = -4