a) Đặt \(t = \sqrt {1 + x} \) , ta được: \(x = t^2- 1, dx = 2t dt\)
Khi \(x = 0\) thì \(t = 1\), khi \(x = 3\) thì \(t = 2.\)
Do đó:
\(\displaystyle \int_0^3 {{x \over {\sqrt {1 + x} }}} dx = \int_1^2 {{{{t^2} - 1} \over t}} .2tdt = 2\int_1^2 {({t^2} - 1)dt}\)
\(\displaystyle= 2({{{t^3}} \over 3} - t)\left| {_1^2} \right. = 2({8 \over 3} - 2 - {1 \over 3} + 1) = {8 \over 3} \)
b) Ta có:
\(\displaystyle\int_1^{64} {{{1 + \sqrt x } \over {\root 3 \of x }}} dx = \int_1^{64} {{{1 + {x^{{1 \over 2}}}} \over {{x^{{1 \over 3}}}}}} dx = \int_1^{64} {({x^{{-1 \over 3}}} + {x^{{1 \over 6}}})dx}\)
\(\displaystyle=({3 \over 2}{x^{{2 \over 3}}} + {6 \over 7}{x^{{7 \over 6}}})\left| {_1^{64}} \right. = \frac{{1872}}{{14}} - \frac{{33}}{{14}}= {{1839} \over {14}}. \)
c) Ta có:
\(\displaystyle \int_0^2 {{x^2}} {e^{3x}}dx = {1 \over 3}\int_0^2 {{x^2}} d{e^{3x}} \) \(\displaystyle = {1 \over 3}{x^2}{e^{3x}}\left| {_0^2} \right. - {2 \over 3}\int_0^2 {x{e^{3x}}} dx \) \(=\dfrac{1}{3}\left. {{x^2}{e^{3x}}} \right|_0^2 - \dfrac{2}{9}\int\limits_0^2 {xd\left( {{e^{3x}}} \right)} \) \(\displaystyle = {4 \over 3}{e^6} - {2 \over 9}(x{e^{3x}})\left| {_0^2} \right. + {2 \over {27}}\int_0^2 {{e^{3x}}} d(3x) \)
\(\displaystyle = {4 \over 3}{e^6} - {4 \over 9}{e^6} + {2 \over {27}}{e^{3x}}\left| {_0^2} \right. = {2 \over {27}}(13{e^6} - 1) \)
d) Ta có:
\( \sqrt {1 + \sin 2x} = \sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x{\mathop{\rm cosx}\nolimits} }\)
\(= |{\mathop{\rm s}\nolimits} {\rm{inx}} + {\mathop{\rm cosx}\nolimits} | \)\(\displaystyle = \sqrt 2 |\sin (x + {\pi \over 4})| \)
\(=\left\{ \matrix{
\sqrt 2 \sin (x + {\pi \over 4}),x \in \left[ {0,{{3\pi } \over 4}} \right] \hfill \cr
- \sqrt 2 \sin (x + {\pi \over 4}),x \in \left[ {{{3\pi } \over 4},\pi } \right] \hfill \cr} \right.\)
Do đó:
\( \displaystyle \int_0^\pi {\sqrt {1 + \sin 2x} } dx = \sqrt 2 \int_0^{{{3\pi } \over 4}} {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4})\) \(\displaystyle - \sqrt 2 \int_{{{3\pi } \over 4}}^\pi {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4}) \) \(\displaystyle = - \sqrt 2 \cos (x + {\pi \over 4})\left| {_0^{{{3\pi } \over 4}}} \right. + \sqrt 2 \cos (x + {\pi \over 4})\left| {_{{{3\pi } \over 4}}^\pi } \right. = 2\sqrt 2 \)
