To evaluate the integral [tex]\( \int t e^{-2t} \, dt \)[/tex], we can use the method of integration by parts. Integration by parts is based on the formula:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
For this integral, we will choose:
[tex]\[ u = t \][/tex]
[tex]\[ dv = e^{-2t} \, dt \][/tex]
First, let's differentiate [tex]\( u \)[/tex] and integrate [tex]\( dv \)[/tex]:
[tex]\[ du = dt \][/tex]
[tex]\[ v = \int e^{-2t} \, dt \][/tex]
To find [tex]\( v \)[/tex], we integrate [tex]\( e^{-2t} \)[/tex]:
[tex]\[ \int e^{-2t} \, dt = \frac{e^{-2t}}{-2} = -\frac{1}{2} e^{-2t} \][/tex]
So, we have:
[tex]\[ v = -\frac{1}{2} e^{-2t} \][/tex]
Now, apply the integration by parts formula:
[tex]\[ \int t e^{-2t} \, dt = t \left(-\frac{1}{2} e^{-2t}\right) - \int \left(-\frac{1}{2} e^{-2t}\right) \, dt \][/tex]
Simplify the first term:
[tex]\[ = -\frac{t}{2} e^{-2t} + \frac{1}{2} \int e^{-2t} \, dt \][/tex]
Next, integrate [tex]\( e^{-2t} \)[/tex] again:
[tex]\[ \int e^{-2t} \, dt = -\frac{1}{2} e^{-2t} \][/tex]
Now substitute this result back into our equation:
[tex]\[ = -\frac{t}{2} e^{-2t} + \frac{1}{2} \left(-\frac{1}{2} e^{-2t}\right) \][/tex]
Simplify the expression:
[tex]\[ = -\frac{t}{2} e^{-2t} - \frac{1}{4} e^{-2t} \][/tex]
Factor out [tex]\( e^{-2t} \)[/tex]:
[tex]\[ = \left(-\frac{t}{2} - \frac{1}{4}\right) e^{-2t} \][/tex]
To express this more cleanly, we combine the coefficients and simplify:
[tex]\[ = \left(\frac{-2t-1}{4}\right) e^{-2t} \][/tex]
Thus, the evaluated integral is:
[tex]\[ \int t e^{-2t} \, dt = \frac{-(2t+1)}{4} e^{-2t} + C \][/tex]
Where [tex]\( C \)[/tex] is the constant of integration.
