\subsection*Solution 4 Let (u=x^2), (du=2x,dx) (\Rightarrow) (x,dx = du/2). When (x=0,u=0); (x=1,u=1). [ \int_0^1 x e^x^2dx = \frac12\int_0^1 e^u du = \frac12(e-1). ]

\subsection*Solution 2 Partition ([0,3]) into (n) equal subintervals: (\Delta x = 3/n), (x_i^* = 3i/n). [ \sum_i=1^n f(x_i^*)\Delta x = \sum_i=1^n \left(2\cdot\frac3in+1\right)\frac3n = \frac3n\left(\frac6n\sum i + \sum 1\right) ] [ = \frac3n\left(\frac6n\cdot\fracn(n+1)2+n\right) = \frac3n\left(3(n+1)+n\right)= \frac3n(4n+3). ] [ \lim_n\to\infty \frac12n+9n = 12. ] Thus (\int_0^3 (2x+1)dx = 12).

Standard Riemann sum definition; continuity ensures integrability.

\subsection*Problem 6 Find the average value of (f(x) = \cos x) on ([0,\pi]).

\subsection*Problem 10 Compute (\int_0^2 \lfloor x \rfloor dx) (greatest integer function).

\section*Basic Problems

Δx = 0.5, right endpoints: 0.5, 1, 1.5, 2. Sum = (0.25 + 1 + 2.25 + 4) × 0.5 = 3.75.

\beginenumerate[label=\arabic*.] \item (\int_0^1 (3x^2-2x+1)dx = 1) \item (\int_1^e \frac1xdx = 1) \item (\int_0^\pi/2 \sin 2x,dx = 1) \item (\int_0^4 |x-2|dx = 4) \item (\lim_n\to\infty \sum_k=1^n \fracnn^2+k^2 = \frac\pi4) \endenumerate