Question

Fan & Medal:)
Can somebody help me with this integration question from Calc? I am lost because the bound goes up to 15.

Answer 1

\[\int_{x=2}^{x=5} g(x) dx = 12\] try and sub u=3x.

Answer 2

\[k\int_{u=?}^{u=?} g(u) du\] for a const. k

Answer 3

*then theres two of you in the question*

Answer 4

I think I'm glitching but you guys have the same avi so I tagged you both haha.

Answer 5

no he stole my picture -.-

Answer 6

c;

Answer 7

wat kind of math is this?

Answer 8

oh calc

Answer 9

calculus

Answer 10

Calc I. I'm assuming you can't help?

Answer 11

@Directrix @inkyvoyd @just_one_last_goodbye @ShadowLegendX @welshfella

Answer 12

hold on bud im tryna get you help

Answer 13

@rishavraj SUMMON

Answer 14

@Michele_Laino Just came online. He can help you ^-^

Answer 15

Thanks! I appreciate it. :)

Answer 16

first step:
we have this
\[\int_2^5 {g\left( x \right)dx} = \int_2^6 {g\left( x \right)dx} + \int_6^5 {g\left( x \right)dx} = 12 - \left( { - 3} \right) = 15\]

Answer 17

if we make this variable change:
\(3x=z\), then we can write:
\[\Large \int_6^{15} {g\left( {3x} \right)dx} = \frac{1}{3}\int_{18}^{45} {g\left( z \right)dz} \]

Answer 18

I'm thinking...

Answer 19

Haha take your time. I really appreciate your help.

Answer 20

Is the ans 15?

Answer 21

No, that's what we get for the integral from 2 to 5.

Answer 22

Oops i meant 5

Answer 23

if i make this variable change:
\[y = 9x\]
I get:
\[\Large 15 = \int_2^5 {g\left( x \right)dx} = \int_{18}^{45} {g\left( {\frac{y}{9}} \right)} \;d\left( {\frac{y}{9}} \right)\]

Answer 24

Sir we can simply do it in this way..
Let us put 3x=t then 3dx=dt
Now what we have to solve if
Integral of g(t)dt/3 within limits 2 and 5..
Well that can be calculated from the data given.. To us..

Answer 25

If I set \(3x=t\) then the interval \([6,15]\) for x, goes to \([15,45]\) for variable t

Answer 26

oops.. I meant \([18,45]\)

Answer 27

Ohh!! Gosh..what a blunder i did here??
So sorry sir...

Answer 28

I'm sorry, I don't know how to continue

Answer 29

i honestly think the problem has a type-o

Answer 30

if we were asked to find
\[\int\limits_{6}^{15} g(\frac{x}{3}) dx\]
this would be possible

Answer 31

It's 3x, not x/3

Answer 32

yes I said I think they made a type-o in the problem
and I was trying to think of a problem that is actually answerable that they could have meant

Answer 33

you cannot answer your question as it is

Answer 34

u = 3dx so dx = 1/3 du
Wouldn't that change the limits of the integral to 2 and 6, therefore leaving the integral from 2 to 6 as the answer?

Answer 35

I'm sorry if I'm annoying if I'm wrong :/

Answer 36

Change the limits to 2 and 5***** ugh sorry

Answer 37

As opposed to the integral going to 18 to 45

Answer 38

yes those limits don't make any sense given what you were given :p
\[\int\limits_6^{15} g(\frac{x}{3}) dx \\ \text{ Let } u=\frac{x}{3} \\ \text{ then } du=\frac{1}{3} dx \\ \text{ and if } x=15 \text{ then } u=\frac{15}{3}=5 \\ \text{ and if } x=6 \text{ then } u=\frac{6}{3}=2 \\ \text{ so } \\ \int\limits_6^{15} g(\frac{x}{3}) dx=\int\limits_2^5 g(u) 3 du = 3 \int\limits_2^5 g(u) du\]
this is answerable where as the integral they asked you to evaluate is not

Answer 39

I'll ask my teacher about it as he seems to think there's a definitive answer.

Answer 40

this is answerable because you can write

to find the integral from 2 to 5 we just need to subtract out the integral from 5 to 6 from 2 to 6

Answer 41

Which is 15.

Answer 42

\[\int\limits_2^6 g(x) dx-\int\limits_5^6 g(x) dx=\int\limits_2^5 g(x) dx\]
but don't forget our integral was being multiplied by something

Answer 43

multiplied by 3 actually

Answer 44

\[3[\int\limits_2^6 g(x)dx-\int\limits_5^6 g(x) dx]=3 \int\limits_2^5 g(x) dx\]

Answer 45

Ohh my, you're right.