Intergration trouble

[Anthonyk2013]

I have to integrate 3e^1/2x dx

The rule I use is e^ax dx becomes 1/a e^ax +c

so 3*(1/ 1/2)e^ 1/2x dx

6e^1/2x+c

Right or wrong?

 

[pka]

I have to integrate 3e^1/2x dx
The rule I use is e^ax dx becomes 1/a e^ax +c
so 3*(1/ 1/2)e^ 1/2x dx
6e^1/2x+c

It is hard to read your notation. If this is what you meant then yes:
$\displaystyle \displaystyle\int {3{e^{\frac{x}{2}}}} dx = 6{e^{\frac{x}{2}}} + c$

 

[Deleted member 4993]

I have to integrate 3e^1/2x dx

The rule I use is e^ax dx becomes 1/a e^ax +c

so 3*(1/ 1/2)e^ 1/2x dx

6e^1/2x+c

Right or wrong?

Three possibilities - following hierarchy of mathematical operations.

3e^1/2x = $\displaystyle \dfrac{3e^1}{2}x$

next possibility...

3e^(1/2 x) = $\displaystyle 3e^{\frac{1}{2}x}$ .......... this the one solved above. Watch those grouping symbols

next possibility

3e^1/(2x) = $\displaystyle 3e^{\frac{1}{2x}}$ ..... This is a horse of a different color - watch those grouping symbols again

Which expression did you want to integrate?

 

[lookagain]

3e^1/(2x) = $\displaystyle 3e^{\frac{1}{2x}}$ ..... This is a horse of a different color - watch those grouping symbols again

Actually, $\displaystyle \$3e^1/(2x) $\displaystyle \$ = $\displaystyle \ \dfrac{3e}{2x}$


3e^(1/(2x)) = $\displaystyle \ 3e^{\frac{1}{2x}}$

 

[Deleted member 4993]

Actually, $\displaystyle \$3e^1/(2x) $\displaystyle \$ = $\displaystyle \ \dfrac{3e}{2x}$


3e^(1/(2x)) = $\displaystyle \ 3e^{\frac{1}{2x}}$

You are correct - I missed those two brackets.