Question

∫_0^4▒(4e^5+ 3x)dx evaluate ..

Answer 1

\[∫_0^4(4e^5+ 3x)dx \]

Answer 2

if you can explain also that would be good

Answer 3

\[4e^5x + \frac{3}{2}x^2 |_0^4\]

Answer 4

how did you get 3/2timesx^2 with the 4/0

Answer 5

x is x^2/2

Answer 6

so do you understand it now

Answer 7

i have no clue what i am doing here thats why i am asking

Answer 8

kk i will review on my own will post few more

Answer 9

how do you evaluate this further

Answer 10

4e5x+32x2|40\[4e^{5}x+32x^{2}|\left(\begin{matrix}4 \\ 0\end{matrix}\right)\]

Answer 11

what is with these binomials?

Answer 12

@av8188 your job is to find a function whose derivative is between the integral sign and the dx
once you do that you evaluate at the upper limit, then evaluate at the lower limit, and then subtract the lower from the upper

Answer 13

also i am going to bet that you meant
\[\int_0^4 (4e^{5x}+3x)dx\] eys?

Answer 14

e^5x no it is not there

Answer 15

so you need to find the "anti derivative of
\[4e^{5x}\] which is
\[\frac{4}{5}e^{5x}\]
and the anti derivative of
\[3x\] which is
\[\frac{3}{2}x^2\]
in other words
\[\int (4e^{3x}+3)dx=\frac{4}{5}e^{5x}+\frac{3}{2}x^2\]

Answer 16

then plug in 4, plug in 0, and subtract

Answer 17

hey
how did you type 4/0

Answer 18

you mean
\[\int_0^4\]

Answer 19

use \int_0^4

Answer 20

no no

Answer 21

but we need to be clear about the first term. i am betting it is
\[4e^{5x}\] and not
\[4e^5\] which one is right?

Answer 22

look at ishans post the end of it has |4/0

Answer 23

no sorry

Answer 24

|_0^4

Answer 25

this thread on the top

Answer 26

that will do it

Answer 27

ah k

Answer 28

so how to solve this now sorry to bug u

Answer 29

underscore gives subscript

Answer 30

happy to do it but let me repeat. what is the first term?

Answer 31

to be honest man i dont know what i am doing here

Answer 32

yes and i can walk you through no problem

Answer 33

but just so we don't get some wrong answer i still need to know what he first term is
is it
\[4e^5\] or is it
\[4e^{5x}\]?

Answer 34

one sec

Answer 35

ok

Answer 36

4e^{5}

Answer 37

\[4e^{5}\]

Answer 38

ok then it is easy. because
\[4e^5\] is a constant

Answer 39

want to go slow or just get answer?

Answer 40

answer is fine on this one i have 2 more so on those you can explain

Answer 41

if answer is all you need it is here
http://www.wolframalpha.com/input/?i=integral+0+to+4+%284e^5%2B3x%29+dx

Answer 42

but this is a very easy one, steps would be simple

Answer 43

please show steps lol

Answer 44

ok here we go

Answer 45

is there a site that shows the steps? save you time...

Answer 46

first of all it is clear that
\[4e^5\] is just a number, a constant, not a function. and you want
\[\int_0^4 4e^5 dx\] meaning the area under the curve (line)
\[y=4e^5\] from x = 0 to x = 4. this is just a rectangle, whose base if 4 (from 0 to 4 is 4) and whose height is
\[4e^5\] so the area of a rectangle is just base times height, we get
\[\int_0^4 4e^5dx=4\times 4e^5=16e^5\]

Answer 47

now we want
\[\int_0^43xdx\] and here you have to find a function whose derivative is 3x
that function is
\[\frac{3}{2}x^2\] and that should be clear once you see it

Answer 48

k

Answer 49

then you can write
\[\frac{3}{2}x^2|_0^4\] if yo like but this just means plug in 4, plug in 0 and subtract. if you plug in 4 you get
\[\frac{3}{2}4^2=\frac{3}{2}\times 16=3\times 8=24\] and if you plug in 0 you get 0

Answer 50

so in total we have
\[\int_0^4 4e^5+3xdx=16e^5+24\] finished

Answer 51

just like worlfram said in the link i sent you

Answer 52

thank you

Answer 53

you probably already know, but I'll show anyway

for \[\frac{3}{2}x^2|_0^4\]
it looks nicer when you do this

\[\left.\frac{3}{2}x^2\right|_0^4\]

\left.\frac{3}{2}x^2\right|_0^4

Answer 54

actually i didn't know. thanks. do i need the left? or just the right?

Answer 55

you need both

Answer 56

\[\frac{3}{2}x^2\right|_0^4\]

Answer 57

yeah i guess so

Answer 58

\[\left.x^3 \right|_0^4\]

Answer 59

oh and i need the dot too. didn't see that one

Answer 60

yes

Answer 61

\[\left.\frac{5}{4}x^3\right|_0^5\]

Answer 62

it just makes the vertical line the same size as you expression

Answer 63

aah it expands to fit the expression i see

Answer 64

thanks!

Answer 65

just like the parentheses

Answer 66

\[\left(\frac{3}{4}\right)\]

Answer 67

\left(\frac{3}{4}\right)

Answer 68

damn mine didn't work

Answer 69

\[\left(\frac{1}{\frac{1}{1+x}}\right)^2\]

Answer 70

\[f(x)=\left\{\begin{matrix}x^2, & x>0 \\ 2-x, & x\leq0\end{matrix}\right.\]
f(x)=\left\{\begin{matrix}x^2, & x>0 \\ 2-x, & x\leq0\end{matrix}\right.

Answer 71

now it did. no dot on this one

Answer 72

wow i was doing this with array!

Answer 73

only use the dot if you don't want to show anything

Answer 74

ooh i see

Answer 75

no screwing upo

Answer 76

sorry that didn't copy correctly

f(x)=\left\{\begin{matrix}x^2, & x>0 \\ 2-x, & x\leq0\end{matrix}\right.


hmmm...this is not displaying correctly...I am using the matrix command
\begin{matrix}

Answer 77

can you give me the whole thing you wrote for piecewise

Answer 78

I'll have to give it in pieces

Answer 79

if you don't mind just copy it in chat

Answer 80

that way i can see it whole because latex does not work in chat

Answer 81

f(x)=\left\{\begin{matrix}x^2, & x>0 \\ 2-x, & x\leq0\end{matrix}\right.

Answer 82

\[f(x)=\left\{\begin{matrix}x^2, & x>0 \\ 2-x, & x\leq0\end{matrix}\right.\]

Answer 83

nice :)

Answer 84

damn damn damn

Answer 85

i went back to chat and it was in latex there too!

Answer 86

latex works in chat when you have question opened matt told me that

Answer 87

do me a favor and send it again

Answer 88

go to group home and then check the latex

Answer 89

thanks

Answer 90

pasted in to text editor

Answer 91

i was using this
f(x) = \left\{
\begin{array}{lr}
1 & : x \in \mathbb{Q}\\
0 & : x \notin \mathbb{Q}
\end{array}
\right.

Answer 92

ic

Answer 93

you know you can see the source code for latex by right clicking on the latex and pressing on 'show source'?

Answer 94

\[f(x)=\left\{\begin{matrix}x^2, & x<0 \\ 2-x, & 0<x<1\\e^x,&x\geq 1\end{matrix}\right.\]

Answer 95

nice :)

Answer 96

man i learn something new every day

Answer 97

and all this time i have been rewriting when i could just copy and past from the code
dope slap for me

Answer 98

\[f(x)=\left\{\begin{matrix}x^2, & x<0 \\ 2-x, & 0<x<1\\e^x,&x\geq 1\end{matrix}\right.\]