Question

**Will give medals** Integral question

Answer 1

for?

Answer 2

\[if \int\limits_{3}^{5} 3f(x) dx=5 and \int\limits_{3}^{5} 2g(x)dx=3, evaluate \int\limits_{3}^{5} 6f(x)+4g(x)+7dx\]

Answer 3

you know you can bring the constants out of the integral ?

\(\int 4 f(y)dy = 4 \int f(y)dy\)
?

Answer 4

yes, but i just dont understand what this question is asking

Answer 5

can you tell me what
\(\int\limits_{3}^{5} f(x) dx=??\)

Answer 6

it doesn't give it

Answer 7

are you talking about the first integral or just in general. Because if your talking about the first integral it is 5

Answer 8

\(\int\limits_{3}^{5} 3f(x) dx=5\)
\(3\int\limits_{3}^{5} f(x) dx=5\)
\(\int\limits_{3}^{5} f(x) dx=5/3\)

does that make sense ?

Answer 9

i understood you took the constant out but how do you get 5/3 from that..

Answer 10

I divided both sides by 3 :)

Answer 11

why though

Answer 12

if your taking the integral wouldnt it be 5x

Answer 13

to get the value of
\(\int\limits_{3}^{5} f(x) dx\)

you'e not given f(x), so you really can't do the actual integration

Answer 14

because if i have
\(\int\limits_{3}^{5} f(x) dx\)

i could just multiply it by 6 to get the value of
\(\int\limits_{3}^{5} 6f(x) dx\)

which I need it to evaluate right side..
doubts?

Answer 15

okay i see

Answer 16

oh sorry im stupid. i see what your saying

Answer 17

there is nothing to integrate. so you divided by 3 to get it to the other side

Answer 18

i understand

Answer 19

right! I'm glad :)
now try get the g(x) part :)

Answer 20

okay

Answer 21

its 3/2 right?

Answer 22

thats
\(\int\limits_{3}^{5} g(x) dx\)
yes.

The question asks for
\(\int\limits_{3}^{5} 4g(x) dx\)

which will be .... ?

Answer 23

4(3/2) ?..

Answer 24

yep

Answer 25

so how do we solve the question completely

Answer 26

\(\large \int\limits_{3}^{5} 6f(x)+4g(x)+7dx \\ \Large = \int\limits_{3}^{5} 6f(x)+\int\limits_{3}^{5} 4g(x)+\int\limits_{3}^{5} 7dx\)

Answer 27

\(\large \int\limits_{3}^{5} 6f(x)+\int\limits_{3}^{5} 4g(x)+\int\limits_{3}^{5} 7dx \\ \Large = 2\times (5/3) + 4\times (3/2) + [7x]^5_3\)

makes sense? can you go further?

Answer 28

isnt the 2 in the second line actually 6?

Answer 29

ofcourse, sorry my typo :P

Answer 30

but do you understand whats going on ?

Answer 31

can you just finish the question? because i dont know how to apply the definite integral numbers inside

Answer 32

yes i do

Answer 33

upper limit - lower limit :)

\(\large \int\limits_{3}^{5} 6f(x)+\int\limits_{3}^{5} 4g(x)+\int\limits_{3}^{5} 7dx \\ \Large = 6\times (5/3) + 4\times (3/2) + [7x]^5_3 \\ \large = 10 +6 + [7\times 5 - 7 \times 3] = ..\)

Answer 34

thank you.

Answer 35

i have another question if you could help :D

Answer 36

welcome ^_^

similar ? can you ask in a different post? so that if I am unavailable, others can help you too :)

Answer 37

sure