Question

\int\limits_6^10( dx/x+2)

Answer 1

hint : \(\large \mathbb{\frac{d}{dx}(\ln (x+2)) = \frac{1}{x+2}}\)

Answer 2

Ok thank you :)

Answer 3

np :) so u can evaluate the integral now ha ?

Answer 4

i tried but i don't know the steps lol

Answer 5

\(\large \mathbb{\int_6^{10} \frac{1}{x+2} dx}\)

\(\large \mathbb{ \ln(x+2) \Big|_2^4}\)

Answer 6

fine so far ?

Answer 7

ok but there is not x it's x+2 and the formula is integral 1/x dx? so there's x+2..

Answer 8

i mean x+2?

Answer 9

integral 1/x+2 dx =ln (x+2?

Answer 10

@ganeshie8

Answer 11

very good question :)

Answer 12

:) i know i am not good at math.little slow actually

Answer 13

\(\large \mathbb{\int \frac{1}{x+2} dx = \ln(x+2) + c}\)

Answer 14

if you differentiate the right side,
wat wud u get ?

Answer 15

nope, u r doing great lol :)

Answer 16

i will get 0? like ln (x+2)/0?

Answer 17

d/dx(2)=0

Answer 18

noo
u need to use chain rule

Answer 19

\(\large \mathbb{\frac{d}{dx} \ln (x+2) = \frac{1}{x+2} * \frac{d}{dx}(x+2)}\)

\(\large \mathbb{~~~~~~~~~~~~~= \frac{1}{x+2} * (1+0)}\)

\(\large \mathbb{~~~~~~~~~~~~~= \frac{1}{x+2} }\)

Answer 20

so we say,

\(\large \mathbb{\int \frac{1}{x+2} = \ln (x+2)}\)

okay ?

Answer 21

oh i got it lol thank u :)

Answer 22

good, so the integral becomes :


\(\large \mathbb{\int_6^{10} \frac{1}{x+2} dx}\)

\(\large \mathbb{ \ln(x+2) \Big|_2^4}\)

Answer 23

if u evaluate the bounds u wud get :

\(\large \mathbb{ \ln(x+2) \Big|_2^4}\)

\(\large \mathbb{ \ln(4+2) - \ln(2+2) }\)

\(\large \mathbb{ \ln(6) - \ln(4) }\)

\(\large \mathbb{ \ln(\frac{6}{4}) }\)

\(\large \mathbb{ \ln(\frac{3}{2}) }\)

Answer 24

see if that makes more or less sense :)

Answer 25

yeah it does lol :) thank u :)

Answer 26

good to hear.... u wlc :)

Answer 27

:)