Bài 23: Cho \(\int\limits_0^1 {f\left( x \right)dx = 3.} \) Tính \(\int\limits_{ – 1}^0 {f\left( x \right)} dx\) trong các trường hợp sau:
a) f là hàm số lẻ; b) f là hàm số chẵn.
Giải
a) f là hàm số lẻ thì \(f\left( { – x} \right) = – f\left( x \right)\)
đặt \(u = – x \Rightarrow du = – dx\)
\(\int\limits_{ – 1}^0 {f\left( x \right)dx} = \int\limits_1^0 {f\left( { – u} \right)\left( { – du} \right)} = \int\limits_0^1 { – f\left( u \right)du}\)
\(= – \int\limits_0^1 {f\left( x \right)dx = – 3.} \)
b) f là hàm số chẵn thì \(f\left( { – x} \right) = f\left( x \right)\)
đặt \(u = – x \Rightarrow du = – dx\)
\(\int\limits_{ – 1}^0 {f\left( x \right)dx = } \int\limits_{ 1}^0 {f\left( { – u} \right)\left( { – du} \right) = } \int\limits_0^1 {f\left( u \right)du = } 3.\)
Bài 24: Tính các tích phân sau :
a) \(\int\limits_1^2 {{x^2}{e^{{x^3}}}dx;} \) b) \(\int\limits_1^3 {{1 \over x}} {\left( {\ln x} \right)^2}dx;\)
c) \(\int\limits_0^{\sqrt 3 } {x\sqrt {1 + {x^2}} } dx;\)
\(d)\,\int\limits_0^1 {{x^2}{e^{3{x^3}}}dx;} \) \(e)\,\int\limits_0^{{\pi \over 2}} {{{\cos x} \over {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}}}} dx.\)
Giải
a) Đặt \(u = {x^3} \Rightarrow du = 3{x^2}dx \Rightarrow {x^2}dx = {{du} \over 3}\)
\(\int\limits_1^2 {{x^2}{e^{{x^3}}}dx = {1 \over 3}} \int\limits_1^8 {{e^u}du = \left. {{1 \over 3}{e^u}} \right|_1^8} = {1 \over 3}\left( {{e^8} – e} \right)\)
b) Đặt \(u = \ln x \Rightarrow du = {{dx} \over x}\)
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\(\int\limits_1^3 {{1 \over x}} {\left( {\ln x} \right)^2}dx = \int\limits_0^{\ln 3} {{u^2}du = \left. {{{{u^3}} \over 3}} \right|} _0^{\ln 3} = {1 \over 3}{\left( {\ln 3} \right)^3}\)
c) Đặt \(u = \sqrt {1 + {x^2}} \Rightarrow {u^2} = 1 + {x^2} \Rightarrow udu = xdx\)
\(\int\limits_0^{\sqrt 3 } {x\sqrt {1 + {x^2}} } dx = \int\limits_1^2 {u.udu = \left. {{{{u^3}} \over 3}} \right|} _1^2 = {7 \over 3}\)
d) Đặt \(u = 3{x^3} \Rightarrow du = 9{x^2}dx \Rightarrow {x^2}dx = {1 \over 9}du\)
\(\int\limits_0^1 {{x^2}{e^{3{x^3}}}dx = {1 \over 9}} \int\limits_0^3 {{e^u}du} = \left. {{1 \over 9}{e^u}} \right|_0^3 = {1 \over 9}\left( {{e^3} – 1} \right)\)
e) Đặt \(u = 1 + {\mathop{\rm s}\nolimits} {\rm{inx}} \Rightarrow du = \cos xdx\)
\(\int\limits_0^{{\pi \over 2}} {{{\cos xdx} \over {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}}}} = \int\limits_1^2 {{{du} \over u}} = \left. {\ln \left| u \right|} \right|_1^2 = \ln 2\)
Bài 25: Tính các tích phân sau :
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a) \(\int\limits_0^{{\pi \over 4}} {x\cos 2xdx;} \) b) \(\int\limits_0^1 {{{\ln \left( {2 – x} \right)} \over {2 – x}}} dx;\)
c) \(\int\limits_0^{{\pi \over 2}} {{x^2}\cos xdx;} \)
\(d)\,\int\limits_0^1 {{x^2}\sqrt {{x^3} + 1} dx;} \) \(e)\,\int\limits_1^e {{x^2}\ln xdx.} \)
Giải
a) Đặt
\(\left\{ \matrix{
u = x \hfill \cr
dv = \cos 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = {1 \over 2}\sin 2x \hfill \cr} \right.\)
Do đó \(\int\limits_0^{{\pi \over 4}} {x\cos 2xdx = \left. {{1 \over 2}x\sin 2x} \right|_0^{{\pi \over 4}}} – {1 \over 2}\int\limits_0^{{\pi \over 4}} {\sin 2xdx} \)
\( = {\pi \over 8} + \left. {{1 \over 4}\cos 2x} \right|_0^{{\pi \over 4}} = {\pi \over 8} + {1 \over 4}\left( { – 1} \right) = {\pi \over 8} – {1 \over 4}.\)
b) Đặt \(u = \ln \left( {2 – x} \right) \Rightarrow du = {{ – 1} \over {2 – x}}dx\)
\(\int\limits_0^1 {{{\ln \left( {2 – x} \right)} \over {2 – x}}} dx = – \int\limits_{\ln 2}^0 {udu} = \int\limits_0^{\ln 2} {udu} \)
\(= \left. {{{{u^2}} \over 2}} \right|_0^{\ln 2} = {1 \over 2}{\left( {\ln 2} \right)^2}\)
c) Đặt
\(\left\{ \matrix{
u = {x^2} \hfill \cr
dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = 2xdx \hfill \cr
v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.\)
Do đó \(I = \int\limits_0^{{\pi \over 2}} {{x^2}\cos xdx = {x^2}} \left. {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^{{\pi \over 2}} – 2\int\limits_0^{{\pi \over 2}} {x\sin xdx}\)
\(= {{{\pi ^2} \over 4}} – 2{I_1}\)
Với \({I_1} = \int\limits_0^{{\pi \over 2}} {x\sin xdx} \)
Đặt
\(\left\{ \matrix{
u = x \hfill \cr
dv = \sin {\rm{x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = – \cos x \hfill \cr} \right.\)
Do đó \({I_1} = – x\left. {\cos x} \right|_0^{{\pi \over 2}} + \int\limits_0^{{\pi \over 2}} {\cos xdx = \left. {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^{{\pi \over 2}}} = 1\)
Vậy \(I = {{{\pi ^2}} \over 4} – 2\)
d) Đặt \(u = \sqrt {{x^3} + 1} \Rightarrow {u^2} = {x^3} + 1 \)
\(\Rightarrow 2udu = 3{x^2}dx \Rightarrow {x^2}dx = {2 \over 3}udu\)
\(\int\limits_0^1 {{x^2}\sqrt {{x^3} + 1} dx} = {2 \over 3}\int\limits_1^{\sqrt 2 } {{u^2}du = \left. {{{2{u^3}} \over 9}} \right|} _1^{\sqrt 2 }\)
\(= {2 \over 9}\left( {2\sqrt 2 – 1} \right)\)
e) Đặt
\(\left\{ \matrix{
u = \ln x \hfill \cr
dv = {x^2}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {{dx} \over x} \hfill \cr
v = {{{x^3}} \over 3} \hfill \cr} \right.\)
Do đó \(\int\limits_1^e {{x^2}\ln xdx = \left. {{{{x^3}} \over 3}\ln x} \right|} _1^e – {1 \over 3}\int\limits_1^e {{x^2}dx}\)
\(= {{{e^3}} \over 3} – \left. {{1 \over 9}{x^3}} \right| _1^e = {{2{e^3} + 1} \over 9}\)
