Bài 4: Sử dụng phương pháp tích phân tưng phần, hãy tính tích phân:
a)\(\int_{0}^{\frac{\pi}{2}}(x+1)sinxdx\) ; b) \(\int_{1}^{e}x^{2}lnxdx\)
c)\(\int_{0}^{1}ln(1+x))dx\) ; d)\(\int_{0}^{1}(x^{2}-2x+1)e^{-x}dx\)
a) Đặt \(u=x+1\); \(dv=sinxdx\) \(\Rightarrow du = dx ;v = -cosx\). Khi đó:
\(\int_{0}^{\frac{\pi}{2}}(x+1)sinxdx=-(x+1)cosx|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}cosxdx\)
\(=1 +sinx|_{0}^{\frac{\pi}{2}}=2\)
b)\(\frac{1}{9}(2e^{3}+1)\). HD: Đặt u = ln x ,dv = x2dx
c) Đặt
\(\eqalign{
& u = \ln x \Rightarrow du = {1 \over x}dx \cr
& dv = {x^2}dx \Rightarrow v = {{{x^3}} \over 3} \cr} \)
Do đó ta có:
\(\int\limits_1^e {{x^2}\ln xdx = {{{x^3}} \over 3}.lnx\left| {_1^e – \int\limits_1^e {{{{x^3}} \over 3}dx = {{{e^3}} \over 3} – \left[ {{{{x^3}} \over 9}} \right]} \left| {_1^e} \right.} \right.}\)\( = {{{e^3}} \over 3} – {{{e^3} – 1} \over 9} = {{2{e^3} + 1} \over 9}\)
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d) Ta có :
\(\int_{0}^{1}(x^{2}-2x-1)e^{x}dx= \int_{0}^{1}(x^{2}-1)e^{-x}dx\)\(-2\int_{0}^{1}x.e^{-x}dx\)
Đặt \(u= {x^2} – 1\); \(dv{\rm{ }} = {\rm{ }}{e^{ – x}}dx\) \(\Rightarrow du = 2xdx ;v = -e^{-x}\) Khi đó :
\(\int_{0}^{1}(x^{2}-1)e^{-x}=-e^{-x}(x^{2}-1)|_{0}^{1}+2\int_{0}^{1}xe^{-x}dx\)
\(=-1+2\int_{0}^{1}x.e^{-x}dx\)
Vậy : \(\int_{0}^{1}(x^{2}-2x+1)e^{-x}dx\) =\(=-1+2\int_{0}^{1}x.e^{-x}dx-2\int_{0}^{1}x.e^{-x}dx\) = -1
Bài 5: Tính các tích phân sau:
a) \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx\) ; b) \(\int_{0}^{\frac{1}{2}}\frac{x^{3}-1}{x^{2}-1}dx\)
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c) \(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx\)
a) \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx =\frac{1}{3}\int_{0}^{1}(1+3x)^{\frac{3}{2}}d(1+3x)\)
\(=\frac{1}{3}\frac{2}{5}(1+3x)^{\frac{5}{2}}|_{0}^{1}=4\tfrac{2}{15}\)
b) \(\int_{0}^{\frac{1}{2}}\frac{x^{3}-1}{x^{2}-1}dx= \int_{0}^{\frac{1}{2}}\frac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)}dx= \int_{0}^{\frac{1}{2}}\frac{x(x+1)+1}{x+1}dx\)
\(=\int_{0}^{\frac{1}{2}}(x+\frac{1}{x+1})dx=(\frac{x^{2}}{2}+ln\left | x+1 \right |)|_{0}^{\frac{1}{2}}=\frac{1}{8}+ln\frac{3}{2}\)
c) Đặt \(u = ln(1+x)\), \(dv=\frac{1}{x^{2}}dx\)\( \Rightarrow du=\frac{1}{1+x},v=-\frac{1}{x}\)
Khi đó :
\(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx = -\frac{1}{x}ln(1+x)|_{1}^{2}+\int_{1}^{2}\frac{dx}{x(1+x)}\)
\(= – {{\ln 3} \over 2} + \ln 2 +\int\limits_1^2 {\left( {{1 \over x} – {1 \over {x + 1}}} \right)dx} \)
\(={\ln {2 \over {\sqrt 3 }} + {\rm{[}}\ln |x| – ln|x + 1|{\rm{]}}\left| {_1^2 = \ln {2 \over {\sqrt 3 }} + \ln {4 \over 3} } \right.}\)
\(= \ln {8 \over {3\sqrt 3 }}\)
Bài 6: Tính tích phân \(\int_{0}^{1}x(1-x)^{5}dx\) bằng hai phương pháp:
a) Đổi biến số : \(u = 1 – x\);
b) Tính tích phân từng phần.
a) Đặt \(u = 1 – x \Rightarrow x = 1 – u \Rightarrow dx = – du\).
Khi \(x = 0\) thì \(u = 1\), khi \(x = 1\) thì \(u = 0\). Khi đó:
\(\int_{0}^{1}x(1-x)^{5}dx=\int_{0}^{1}(1-u)u^{5}du=(\frac{1}{6}u^{6}-\frac{1}{7}u^{7})|_{0}^{1}\)\(=\frac{1}{42}.\)
b) Đặt \(u = x\); \(dv = (1 – x)^5dx\)
\(\Rightarrow du = dx\); \(v=-\frac{(1+x)^{6}}{6}\). Khi đó:
\(\int_{0}^{1}x(1-x)^{5}dx=-\frac{x(1-x)^{6}}{6}|_{0}^{1}+\frac{1}{6}\int_{0}^{1}(1-x)^{6}dx\)
\(=-\frac{1}{6}\int_{0}^{1}(1-x)^{6}d(1-x)=-\frac{1}{42}(1-x)^{7}|_{0}^{1}=\frac{1}{42}\).
