Question

Definite Integrals

Answer 1

Suppose \[\Large \int\limits_{2}^{8}g(x) dx=5\]
and
\[\Large \int\limits_{6}^{8} g(x) dx= -3\]
Find the value of \[\Large \int\limits_{4}^{12} g(2x) dx \]

Answer 2

I got \[\Large \int\limits_{2}^{6}g(x)dx+\int\limits_{6}^{8}g(x)dx=5\]
\[\Large \int\limits\limits_{2}^{6}g(x)dx=5-\int\limits\limits_{6}^{8}g(x)dx\]
\[\Large \int\limits\limits_{2}^{6}g(x)dx=5-(-3)=8\]

Answer 3

@campbell_st or @Compassionate can I have some help?

Answer 4

u are absolutely correct

Answer 5

Ok, where do I go from there

Answer 6

8 is the answer

Answer 7

promise me a medal...:P

Answer 8

I'm trying to find \[\Large \int\limits\limits_{4}^{12} g(2x) dx\], all of that up there was what i could get from the question that I know of

Answer 9

are you acquianted with the properties of definite integral?

Answer 10

Yeah

Answer 11

I should add the answer choices are

8

12

16

4

Answer 12

take 2x=k..dk=2dx...
now the integral looks like g(k)dk*1/2..
u will need to change limits too...the limits will be reduced by half
so essentially ur integral becomes limit 2,6 (g(k)*dk*.5)..now change the variable to x
u have limit (2,6) (g(x)dx*.5)
thats 4..oops i was wrong..its half of 8

Answer 13

get it?

Answer 14

Hold on, I'm working it through

Answer 15

Wouldn't the limits be doubled?

Answer 16

@amriju

Answer 17

double

Answer 18

pellet

Answer 19

m an retriceole

Answer 20

Lol, it's ok, in that case that they are doubled it would now look like
\[\Large \frac{1}{2} \cdot\int\limits\limits_{8}^{24}g(k) dk\]

Answer 21

take 2k=x...put it in the integral limit (2,6)

Answer 22

How come 2k=x?

Answer 23

i think now it should come

Answer 24

take that assumption and substitute x by k in limit(2,6) (g(x)dx)

Answer 25

x by 2k

Answer 26

is kf(x) the same as f(kx)?

Answer 27

No, cause kf(x) all the y values are multiplied by a factor of k while f(kx) changes the x values by the value of k and then it's evaluated in the function

Answer 28

kf(x) deals with vertical stretches/ compression while f(kx) deals with horizontal stretches and compression

Answer 29

correct
so why did you use 1/2 as a constant after doubling the limits of integration?

http://finedrafts.com/files/CUNY/math/calculus/Spivak/Michael%20Spivak%20-%20Calculus.pdf

Answer 30

What page would I be turning to?

Answer 31

*scrolling

Answer 32

I did the k-substitution of k=2x
then dk=2 dx
dk/2=dx and then to find the upper and lower limits if we keep g(k) is

(12) upper-> 2(12)=24
(4) lower-> 2(4)=8

The limits get doubled there and we plug all that in

Answer 33

provisional definition of a function in chapter 3 then jump to definite integral

Answer 34

that is too early use of a substitution, is it even required?

Answer 35

I'm not sure, i was just going along with the other person

Answer 36

laughing out loud

Answer 37

Help :0, I did a whole bunch of substituiton problems in this lesson then they ask me this question that I didn't see before

Answer 38

did u do it?

Answer 39

Yeah, but I'm being told it's not right, can you help @ganeshie8

Answer 40

why cant u substitute x by 2k?

Answer 41

Which part, before or after I did the sub

Answer 42

x=2k...so dx=2dk
g(x)dx=2g(2k)dk..limit will change to double...
now u know

Answer 43

im getting this :
\[\Large \int\limits\limits_{1}^{3} g(2x) dx = 8\]

Answer 44

for x<2 and x>8, we don't even know how the function g(x) behaves, so we cannot calculate the area between 8 and 24 for g(x).

Answer 45

Ok, so what's next

Answer 46

And I thought \[\Large \int\limits\limits\limits_{2}^{6} g(2x) dx = 8\]