Bài 5 trang 127 SGK Giải tích 12: Ôn tập Chương 3 - Nguyên hàm...

Bài 5. Tính:

a) $\int_0^3 {{x \over {\sqrt {1 + x} }}} dx$

b) $\int_1^{64} {{{1 + \sqrt x } \over {\root 3 \of x }}} dx$

c) $\int_0^2 {{x^2}} {e^{3x}}dx$

d) $\int_0^\pi  {\sqrt {1 + \sin 2x} } dx$

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:

$\int_0^3 {{x \over {\sqrt {1 + x} }}} dx = \int_1^2 {{{{t^2} - 1} \over t}} .2tdt = 2\int_1^2 {({t^2} - 1)dt}$

$= 2({{{t^3}} \over 3} - t)\left| {_1^2} \right. = 2({8 \over 3} - 2 - {1 \over 3} + 1) = {8 \over 3}$

b)

Ta có:

$\int_1^{64} {{{1 + \sqrt x } \over {\root 3 \of x }}} dx = \int_1^{64} {{{1 + {x^{{1 \over 2}}}} \over {{x^{{1 \over 3}}}}}} dx = \int_1^{64} {({x^{{-1 \over 3}}} + {x^{{1 \over 6}}})dx}$
$=({3 \over 2}{x^{{2 \over 3}}} + {6 \over 7}{x^{{7 \over 6}}})\left| {_1^{64}} \right. = {{1839} \over {14}}$

c) Ta có:

$\int_0^2 {{x^2}} {e^{3x}}dx = {1 \over 3}\int_0^2 {{x^2}} d{e^{3x}} = {1 \over 3}{x^2}{e^{3x}}\left| {_0^2} \right.$

$- {2 \over 3}\int_0^2 {x{e^{3x}}} dx$$= {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)$
$= {4 \over 3}{e^6} - {4 \over 9}{e^6} + {2 \over {27}}{e^{3x}}\left| {_0^2} \right. = {2 \over {27}}(13{e^6} - 1)$

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} |$$= \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:

$\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})$$- \sqrt 2 \int_{{{3\pi } \over 4}}^\pi {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4})$  $= - \sqrt 2 \cos (x + {\pi \over 4})\left| {_0^{{{3\pi } \over 4}}} \right. + \sqrt 2 (x + {\pi \over 4})\left| {_{{{3\pi } \over 4}}^\pi } \right. = 2\sqrt 2$