Bài 5 trang 127 SGK Giải tích 12
Bài 5 trang 127 SGK Giải tích 12 Tính: ...
Bài 5 trang 127 SGK Giải tích 12
Tính:
Bài 5. Tính:
a) (int_0^3 {{x over {sqrt {1 + x} }}} dx)
b) (int_1^{64} {{{1 + sqrt x } over { oot 3 of x }}} dx)
c) (int_0^2 {{x^2}} {e^{3x}}dx)
d) (int_0^pi {sqrt {1 + sin 2x} } dx)
Trả lời:
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 = 2int_1^2 {({t^2} - 1)dt})
(= 2({{{t^3}} over 3} - t)left| {_1^2} ight. = 2({8 over 3} - 2 - {1 over 3} + 1) = {8 over 3} )
b)
Ta có:
(int_1^{64} {{{1 + sqrt x } over {
oot 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}}
ight. = {{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} ight.)
(- {2 over 3}int_0^2 {x{e^{3x}}} dx )(= {4 over 3}{e^6} - {2 over 9}(x{e^{3x}})left| {_0^2}
ight. + {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}
ight. = {2 over {27}}(13{e^6} - 1) )
d)
Ta có:
( sqrt {1 + sin 2x} = sqrt {{{sin }^2}x + {{cos }^2}x + 2sin x{mathop{ m cosx} olimits} })
(= |{mathop{ m s} olimits} { m{inx}} + {mathop{ m cosx} olimits} | )(= sqrt 2 |sin (x + {pi over 4})| )
(=left{ matrix{
sqrt 2 sin (x + {pi over 4}),x in left[ {0,{{3pi } over 4}}
ight] hfill cr
sqrt 2 sin (x + {pi over 4}),X in left[ {{{3pi } over 4},pi }
ight] hfill cr}
ight.)
Do đó:
( int_0^pi {sqrt {1 + sin 2x} } dx = sqrt 2 int_0^{{{3pi } over 4}} {sin (x + {pi over 4}} )d(x + {pi over 4}))( - sqrt 2 int_{{{3pi } over 4}}^pi {sin (x + {pi over 4}} )d(x + {pi over 4}) ) (= - sqrt 2 cos (x + {pi over 4})left| {_0^{{{3pi } over 4}}} ight. + sqrt 2 (x + {pi over 4})left| {_{{{3pi } over 4}}^pi } ight. = 2sqrt 2 )
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