HELP! Can someone integrate these? (1 Viewer)

[tooheyz]

Originally posted by McLake
2. find intergral S 3^x dx

let y = 3^x
log3y = x
use chang of base
log3y = ln y/ ln3
lny = xln3
y = e^(xln3)
dy/dx = ln3*e^(xln3)

ah mate im confused

umm the answer is (1/In3)*3^x

 

[tooheyz]

ah yeah no worries man... i figured the other part out... LOL i mistook to 'x' for a miltiply .. i dont know why but yeah...

thanks again

 

[elizabethy]

Originally posted by ToOhEyZ
yeah thats right. i got the first part right as well
after like an hour of analysing it ..

well i've been trying to get the second part right.. i cant get it!

when u intergrate xe^x2 between 1 and 0, it should be 1/2 (e-1) and should be approx 0.86.

yeah thats the answer and i cant get it!

okey so the derivative of y=e^x2
is 2xe^x2

and the integral of xe^x2 from 0 to 1 is
0.5e^x2
evaluate it with a calculator and u will get it 0.859, round it to 0.86

 

[Harimau]

Originally posted by ToOhEyZ


ah mate im confused

umm the answer is (1/In3)*3^x

d/dx (3^x) = ln3*3^x
(1/ln3).d/dx(3^x) = 3^x (1)

Now you want S (3^x) dx.

Therefore you intergrate both sides of (1)

S [(1/ln3).d/dx(3^x)] dx = S (3^x) dx

Now RHS is what you want, and on the LHS the intergral sign and the derivative sign will cancel each other out.

(Therefore) S (3^x) dx = (1/ln3)(3^x)

This is called Intergration by Dedduction.

Another way of doing this, (much shorter)

is to get a function and its derivative inside [ S (3^x) dx ]

Now d/dx (3^x) = ln3*3^x

So you want S ln3.(3^x) dx to balance, and since ln3 is a constant, you get (1/ln3).S ln3.(3^x)

Then you just do it as normal and you get (1/ln3)*(3^x)

Get it?