Question

this is really confusing questions.
I know i have to use subsutition rules to solve.. but how?

Answer 1

@amistre64

Answer 2

\(\int \limits_0^{0.5}f(2t) dt= \int \limits_0^{0.5}f(2u) du \\put \quad 2u=t, \\ =(1/2)\int \limits_0^1f(t)dt\)
which step u have doubt with ?

Answer 3

I think I got it. So the (1.2) is the answer?
once i got it, i will try next question bymyself!

Answer 4

1.2 ?
how ?

Answer 5

I mean 1/2 S f(t) dt

Answer 6

yeah and you know the value of integral f(t) dt , right ?

Answer 7

its 15, right ?
so, A is 15/2

Answer 8

I got 12...
How did you get 15??

Answer 9

\(\int \limits_0^{0.5}f(2t) dt= \int \limits_0^{0.5}f(2u) du \\put \quad 2u=t, \\ =(1/2)\int \limits_0^1f(t)dt=(1/2)\times 15= 15/2\)

Answer 10

how come you got 15?

Answer 11

15*1/2?

Answer 12

why do you keep on asking how i got 15 ?
isn't 15 the value of the original integral and iven in the question ?

Answer 13

*given

Answer 14

\(\ \int \limits_0^1f(t)dt=15\)
given

Answer 15

OHHH sorry. i got it!

Answer 16

ok, no problem....

Answer 17

for the second one., i got

Answer 18

\[\int\limits_{0}^{0.25}f(1-4t)dt=\int\limits_{0}^{0.25}(4u)du?\]

Answer 19

you can put
1-4u =t

Answer 20

\[\int\limits_{0}^{0.25}(1-4t)dt?\]

Answer 21

@hartnn

Answer 22

thats your Q.
first you can change the variable, like i did in part A
\(\int\limits_{0}^{0.25}(1-4t)dt =\int\limits_{0}^{0.25}(1-4u)du\)
now put 1-4u =t
dt=... ?

Answer 23

or du= ... ?

Answer 24

dt=4dx?

Answer 25

dt = 4du or -4du ?

Answer 26

-4du!

Answer 27

yes, now change the limits also, can you ?

Answer 28

and also put du =(-1/4) dt

Answer 29

how do you limit?
you mean i have to set the equation?

Answer 30

find the upper and lower limits
when u = 0, t =... ?
when u = 0.25, t= ... ?

Answer 31

t=1
t=0?

Answer 32

yes.

Answer 33

\(\int\limits_{0}^{0.25}(1-4t)dt =\int\limits_{0}^{0.25}(1-4u)du= \int \limits_1^0f(t)(-1/4)dt\)
got this ?

Answer 34

so far yes

Answer 35

now, you can use,
\(\int \limits_a^bf(x)dx=-\int \limits_b^af(x)dx\)

Answer 36

so,
\(\large \int\limits_{0}^{0.25}(1-4t)dt=\int \limits_1^0f(t)(-1/4)dt=\int \limits_0^1f(t)(1/4)dt\)
ok ?

Answer 37

yes,

Answer 38

the next step is... 1/4 *15?

Answer 39

good!

Answer 40

now can you try part C ?

Answer 41

I will ! i will write my equation first.

Answer 42

I entered B as 4/15, and tells me that it was wrong

Answer 43

No sorry it was right...

Answer 44

for C is it 15*1/10?
I just guessed

Answer 45

4-10 u = t
dt = -10 du

Answer 46

I see!

Answer 47

so, yes its 15/10 or 3/2

Answer 48

Thank you very much for being helpful!

Answer 49

hey, you are always welcome ^_^