Question

Please Help!

Answer 1

step to step.....

Answer 2

\[\huge \int\limits_{1}^{4}\frac{3x^3-2x^2+4}{x^2} dx\]
Next divide throught by x^2 we find:
\[\huge \rightarrow \int\limits_{1}^{4}[\frac{3x^3}{x^2}-\frac{2x^2}{x^2}+\frac{4}{x^2} ]dx\]
\[\huge \rightarrow \int\limits_{1}^{4}[3x-2+4x^{-2} ]dx\]

\[\huge \rightarrow 3\int\limits_{1}^{4}x dx- \int\limits_{1}^{4}2+4 \int\limits_{1}^{4}x^{-2} dx\]
\[\huge \rightarrow 3\int\limits_{1}^{4}x dx- 2\int\limits_{1}^{4}1dx+4 \int\limits_{1}^{4}x^{-2} dx\]
\[\huge \rightarrow 3\frac{x^2}{2}- 2x+4 \frac{x^{-1}}{-1}\]
\[\huge \rightarrow \frac{3x^2}{2}- 2x-4 x^{-1}\]
\[\huge \rightarrow \frac{3x^2}{2}- 2x-\frac{4}{x}\]
is the required answer of the given integral

@fabiomartins

Answer 3

Oh sorry, i just forgot to apply the lmits

Answer 4

the end result is this?

Answer 5

\[\huge \frac{3}{2}[x^2]_1^4−2[x]_1^4−4[\frac{1}{x}]_1^4\]
\[\huge \rightarrow \frac{3}{2}[16-1]_1^4−2[4-1]_1^4−4[\frac{1}{4-1}]_1^4\]
\[\huge \rightarrow \frac{3}{2}[15]−2[3]−4[\frac{1}{3}]\]
\[\huge \rightarrow \frac{45}{2}−6−\frac{4}{3}\]
\[\huge \rightarrow \frac{135}{6}−\frac{36}{6}−\frac{8}{6}\]
\[\huge \rightarrow \frac{135-36-8}{6} =\frac{135-44}{6} =\frac{91}{6} \]
This is the end result
@fabiomartins

Answer 6

I found another friend result ...

Answer 7

Int(3x^3/x^2) - int(2x^2/x^2) + int(4/x^2)
=int(3x) - int(2) + int(4/x^2)
=[1.5x^2] - [2x] + [-4/x]
=(1.5*16-1.5)-(2*4-2)-(4/4-4/1)
=22.5-6+3
= 19,5

Answer 8

the result has to give 19.5

Answer 9

In the way you wrote it, you forgot to square the limits when you plugged it into the x^2 terms of the antiderivative.

Answer 10

Dpasingh is correct.

Answer 11

@dpasingh \[[\frac{ 1 }{ x }]^{4}_{1} = [\frac{ 1 }{ 4 } - 1]\]

Answer 12

look at the result in the site

Answer 13

No it would be \[\frac{1}{4-1}\]

Answer 14

http://www.wolframalpha.com/widgets/view.jsp?id=1baf76735e53fba2b533ad071288344

Answer 15

Result

Answer 16

when you replace the boundaries , you must replace for the whole function not only x

example

\[[3x]^2_1 = [ 3(2) - 3(1)] = 3[2-1]\]

and same for fractions

Answer 17

@fabiomartins
19.5 is correct

Answer 18

but do you understand how your friend did it ?

Answer 19

more or less, how would you do it your way step by step?

Answer 20

same steps as dpasingh

just where there is \[4[\frac{ 1 }{ x }]^4_1 = 4[\frac{ 1 }{ 4-1 }]\]

replace it by \[4[\frac{ 1 }{ x }]^4_1 = 4[\frac{ 1 }{ 4 } -1] \]

Answer 21

so you will get

\[\frac{ 3 }{ 2 }[15] - 2[3] -4[\frac{ -3 }{ 4 }]\]

then simplify

Answer 22

makes the final part please?

Answer 23

just multiply, then add


\[\frac{ 45 }{ 2 } - 6 +3\]

can you continue now ?

Answer 24

19,5

Answer 25

thank you friend

Answer 26

you have facebook?

Answer 27

any time :)

Answer 28

yeah lol

Answer 29

me add?

Answer 30

https://www.facebook.com/eng.fabiomartins