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Tính:

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

Trả lời:

a) Đặt \(t = \sqrt {1 + x} \) , ta được: x = t2 – 1, dx = 2t dt

Khi x = 0 thì t = 1, khi x = 3 thì t = 2

Do đó: 

\(\eqalign{
& \int_0^3 {{x \over {\sqrt {1 + x} }}} dx = \int_1^2 {{{{t^2} - 1} \over t}} .2tdt = 2\int_1^2 {({t^2} - 1)dt} \cr
& = 2({{{t^3}} \over 3} - t)\left| {_1^2} \right. = 2({8 \over 3} - 2 - {1 \over 3} + 1) = {8 \over 3} \cr} \)

b)

Ta có:

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

c) Ta có:

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

d)

Ta có:

\(\eqalign{
& \sqrt {1 + \sin 2x} = \sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x{\mathop{\rm cosx}\nolimits} } = |{\mathop{\rm s}\nolimits} {\rm{inx}} + {\mathop{\rm cosx}\nolimits} | \cr
& = \sqrt 2 |\sin (x + {\pi \over 4})| = \left\{ \matrix{
\sqrt 2 \sin (x + {\pi \over 4}),x \in \left[ {0,{{3\pi } \over 4}} \right] \hfill \cr
\sqrt 2 \sin (x + {\pi \over 4}),X \in \left[ {{{3\pi } \over 4},\pi } \right] \hfill \cr} \right. \cr} \)

Do đó: 

\(\eqalign{
& \int_0^\pi {\sqrt {1 + \sin 2x} } dx = \sqrt 2 \int_0^{{{3\pi } \over 4}} {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4}) - \sqrt 2 \int_{{{3\pi } \over 4}}^\pi {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4}) \cr
& = - \sqrt 2 \cos (x + {\pi \over 4})\left| {_0^{{{3\pi } \over 4}}} \right. + \sqrt 2 (x + {\pi \over 4})\left| {_{{{3\pi } \over 4}}^\pi } \right. = 2\sqrt 2 \cr} \)