Solving the Integral: $\int_{0}^{\ln 2} e^{-2x} \,dx$

[tandoorichicken]

how do i do
\int_{0}^{\ln 2} e^{-2x} \,dx
?

 

[cookiemonster]

Try a substitution of u = -2x. Don't forget to convert the boundaries of integration.

cookiemonster

 

[himanshu121]

Formula : \int e^{ax}dx= \frac{e^{ax}}{a}+c

 

[cookiemonster]

Figured since himanshu threw out a formula from nowhere, I'd go ahead and derive it. I know I always hated formulas from nowhere.

\int_a^b e^{\alpha x}\,dx
make the substitution u = \alpha x. Therefore, du = \alpha dx.
\int_a^b e^{\alpha x}\,dx = \int_{a'}^{b'} e^{u}\frac{du}{\alpha} = \frac{1}{\alpha} \int_{a'}^{b'} e^{u}\,du
\frac{1}{\alpha} \int_{a'}^{b'} e^u\,du = \frac{1}{\alpha} e^u \Big|^{b'}_{a'} = \frac{1}{\alpha} e^{\alpha x} \Big|^b_a = \frac{1}{\alpha}(e^{\alpha b} - e^{\alpha a})

cookiemonster

 

[cookiemonster]

Just a note: The constant of integration is superfluous in a definite integral, such as this problem. It just cancels itself out at the end anyway.

cookiemonster

 

[himanshu121]

Firstly i wrote it in the form of indefinite integral so from the rul;e i wrote c now u c y i wrote c

it is not from nowhere
\frac{d e^{ax}}{dx}= ae^{ax}

Now rearrange u will get

\int e^{ax}dx= \frac{e^{ax}}{a}+c