Tính
LG a
\(\int\limits_0^1 {\sqrt {{t^5} + 2t} } \left( {2 + 5{t^4}} \right)dt;\)
Lời giải chi tiết:
Đặt \(u = \sqrt {{t^5} + 2t} \Rightarrow {u^2} = {t^5} + 2t \) \(\Rightarrow 2udu = \left( {5{t^4} + 2} \right)dt\)
| t | 0 | 1 |
| u | 0 | \(\sqrt 3 \) |
\(\int\limits_0^1 {\sqrt {{t^5} + 2t} } \left( {2 + 5{t^4}} \right)dt = \int\limits_0^{\sqrt 3 } {2{u^2}du} \) \( = \left. {{{2{u^3}} \over 3}} \right| _0^{\sqrt 3 } \) \( = \dfrac{{2{{\left( {\sqrt 3 } \right)}^3}}}{3} - 0= 2\sqrt 3 \)
LG b
\(\int\limits_0^{{\pi \over 2}} {x\sin {\rm{xcosx}}dx} .\)
Lời giải chi tiết:
Ta có \(\displaystyle I = \int\limits_0^{{\pi \over 2}} {x\sin x\cos xdx} \) \(\displaystyle = {1 \over 2} \int\limits_0^{{\pi \over 2}} {x\sin 2xdx} \)
Đặt
\(\displaystyle \left\{ \matrix{
u = x \hfill \cr
dv = \sin 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = - {1 \over 2}\cos 2x \hfill \cr} \right.\)
Do đó \(\displaystyle I = \left. {{1 \over 2}\left( { - {1 \over 2}x\cos 2x} \right)} \right|_0^{{\pi \over 2}} + {1 \over 4}\int\limits_0^{{\pi \over 2}} {\cos 2xdx }\) \(\displaystyle = \frac{1}{2}\left( { - \frac{1}{2}.\frac{\pi }{2}\cos \pi - 0} \right) + \frac{1}{4}.\left. {\frac{1}{2}\sin 2x} \right|_0^{\frac{\pi }{2}}\) \(\displaystyle = {\pi \over 8} + \left. {{1 \over 8}\sin 2x} \right|_0^{{\pi \over 2}} \) \(\displaystyle = \frac{\pi }{8} + \frac{1}{8}\left( {\sin \pi - \sin 0} \right)= {\pi \over 8}\)
