int3e^(3t)dt?

Feb 19, 2018

Assuming $e$ to be Euler's constant and $t$ our variable,

$\int 3 {e}^{3 t} \mathrm{dt} = {e}^{3 t} + c , c \in \mathbb{R}$

Explanation:

When you take an indefinite integral of an expression (a function of a variable) with respect to a variable ($t$ in our case), the result is (the general form of all) functions which when differentiated with respect to that variable give us our initial function (what we have in the integral). So we can check our answer by differentiating

${e}^{3 t} + c$

with respect to $t$, for any real constant $c$. The reason we put the constant there is because its derivative is $0$ no matter its value, giving us a general form for all possible functions that fit the description above.

$\left({e}^{3 t} + c\right) ' = \left(3 t\right) ' {e}^{3 t} + \left(c\right) ' = 3 {e}^{3 t} + 0 = 3 {e}^{3 t}$

Verifying our solution.