부분적분법

(a) By the Second Fundamental Theorem of Calculus, $F'(x) = f(x) = xe^{-2x}$. (b) To use the First Fundamental Theorem of Calculus to evaluate the integral in $F(x) = \int_0^x te^{-2t} \, dt$, we use integration by parts. Letting $$\begin{aligned}u &= t, \ du = dt \\ dv &= e^{-2t} \, dt, \ v = -\frac{1}{2}e^{-2t}\end{aligned}\,,$$ it follows that $$\int te^{-2t} \, dt = -\frac{1}{2}te^{-2t} - \int \left( -\frac{1}{2}e^{-2t} \right) \, dt = -\frac{1}{2}te^{-2t} - \frac{1}{4}e^{-2t}\,.$$ Thus, $$\begin{aligned}F(x) &= \int_0^x te^{-2t} \, dt \\ &= \left. -\frac{1}{2}te^{-2t} - \frac{1}{4}e^{-2t} \right|_{0}^{x} \\ &= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} - \left( 0 - \frac{1}{4}e^{0} \right) \\ &= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + \frac{1}{4}\end{aligned}$$ (c) For $x > 0$, we observe that $F'(x) = f(x) = xe^{-2x} > 0$ since $e$ to any power is positive. Thus, because $F'(x) > 0$, it follows that $F$ is increasing for all $x > 0$.