I=1∫0(xe2x−x√4−x2)dx=1∫0xe2xdx−1∫0x√4−x2dx=J−K. ★J=1∫0xe2xdx
Đặt {u=xdv=e2xdx, khi đó {du=dxv=12e2x, ta có:
J=12xe2x|10−1∫012e2xdx=e22−14e2x|10=e22−14(e2−1)=e24+14
★K=1∫0x√4−x2dx
Đặt t=√4−x2⇒t2=4−x2⇒2tdt=−2xdx⇔tdt=−xdx, ta có:
x=0⇒t=2;x=1⇒t=√3
K=2∫√3ttdt=t|2√3=2−√3
Suy ra: I=(e24+14)−(2−√3)=e24+√3−74.