Câu hỏi/bài tập:
Tính
a) \(\int\limits_{ - 1}^1 {4{x^7}dx} \)
b) \(\int\limits_{ - 2}^{ - 1} {\frac{{ - 3}}{{10x}}dx} \)
c) \(\int\limits_0^2 {\frac{{{5^{x - 1}}}}{2}dx} \)
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Sử dụng tính chất \(\int\limits_a^b {kf\left( x \right)dx} = k\int\limits_a^b {f\left( x \right)dx} \) để tính các tích phân.
a) \(\int\limits_{ - 1}^1 {4{x^7}dx} = 4\int\limits_{ - 1}^1 {{x^7}dx} = 4\left. {\left( {\frac{{{x^8}}}{8}} \right)} \right|_{ - 1}^1 = 4\left[ {\frac{{{1^8}}}{8} - \frac{{{{\left( { - 1} \right)}^8}}}{8}} \right] = 0\).
b) \(\int\limits_{ - 2}^{ - 1} {\frac{{ - 3}}{{10x}}dx} = \frac{{ - 3}}{{10}}\int\limits_{ - 2}^{ - 1} {\frac{1}{x}dx} = \frac{{ - 3}}{{10}}\left. {\left( {\ln \left| x \right|} \right)} \right|_{ - 2}^{ - 1} = \frac{{ - 3}}{{10}}\left( {\ln \left| { - 1} \right| - \ln \left| { - 2} \right|} \right) = \frac{{3\ln 2}}{{10}}\)
c) \(\int\limits_0^2 {\frac{{{5^{x - 1}}}}{2}dx} = \int\limits_0^2 {\frac{{{5^x}}}{{2.5}}dx} = \frac{1}{{10}}\int\limits_0^2 {{5^x}dx} = \frac{1}{{10}}.\left. {\left( {\frac{{{5^x}}}{{\ln 5}}} \right)} \right|_0^2 = \frac{1}{{10}}\left( {\frac{{{5^2}}}{{\ln 5}} - \frac{{{5^0}}}{{\ln 5}}} \right) = \frac{{12}}{{5\ln 5}}\)