Câu hỏi/bài tập:
Tính
a) \(\int\limits_{ - 1}^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} - \int\limits_1^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} \)
b) \(\int\limits_0^3 {\left| {x - 1} \right|dx} \)
c) \(\int\limits_0^\pi {\left| {\cos x} \right|dx} \)
a) Sử dụng các tính chất của tích phân: \(\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} \) và \(\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} \).
b) Ta có \(\left| {x - 1} \right| = \left\{ {\begin{array}{*{20}{c}}{x - 1{\rm{ }}\left( {x \ge 1} \right)}\\{1 - x{\rm{ }}\left( {x < 1} \right)}\end{array}} \right.\). Từ đó ta có \(\int\limits_0^3 {\left| {x - 1} \right|dx} = \int\limits_0^1 {\left| {x - 1} \right|dx} + \int\limits_1^3 {\left| {x - 1} \right|dx} \)
c) Ta có \(\left| {\cos x} \right| = \left\{ {\begin{array}{*{20}{c}}{\cos x{\rm{ }}\left( {0 \le x \le \frac{\pi }{2}} \right)}\\{ - \cos x{\rm{ }}\left( {\frac{\pi }{2} \le x \le \frac{\pi }{2}} \right)}\end{array}} \right.\).
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Từ đó ta có \(\int\limits_0^\pi {\left| {\cos x} \right|dx} = \int\limits_0^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} + \int\limits_{\frac{\pi }{2}}^\pi {\left| {\cos x} \right|dx} \).
a) \(\int\limits_{ - 1}^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} - \int\limits_1^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} = \int\limits_{ - 1}^{\frac{1}{2}} {\left( {4{x^3} - 5} \right)dx} + \int\limits_{\frac{1}{2}}^1 {\left( {4{x^3} - 5} \right)dx} = \int\limits_{ - 1}^1 {\left( {4{x^3} - 5} \right)dx} \)
\( = 4\int\limits_{ - 1}^1 {{x^3}dx} - 5\int\limits_{ - 1}^1 {dx} = \left. {\left( {{x^4}} \right)} \right|_{ - 1}^1 - 5\left. {\left( x \right)} \right|_{ - 1}^1 = \left[ {{1^4} - {{\left( { - 1} \right)}^4}} \right] - 5\left[ {1 - \left( { - 1} \right)} \right] = - 10\)
b) \(\int\limits_0^3 {\left| {x - 1} \right|dx} = \int\limits_0^1 {\left| {x - 1} \right|dx} + \int\limits_1^3 {\left| {x - 1} \right|dx} = \int\limits_0^1 {\left( {1 - x} \right)dx} + \int\limits_1^3 {\left( {x - 1} \right)dx} = \left. {\left( {x - \frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^3\)
\( = \left[ {\left( {1 - \frac{{{1^2}}}{2}} \right) - \left( {0 - \frac{{{0^2}}}{2}} \right)} \right] + \left[ {\left( {\frac{{{3^2}}}{2} - 3} \right) - \left( {\frac{{{1^2}}}{2} - 1} \right)} \right] = \frac{1}{2} + 2 = \frac{5}{2}\)
c) \(\int\limits_0^\pi {\left| {\cos x} \right|dx} = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} + \int\limits_{\frac{\pi }{2}}^\pi {\left( { - \cos x} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} - \int\limits_{\frac{\pi }{2}}^\pi {\cos xdx} = \left. {\left( {\sin x} \right)} \right|_0^{\frac{\pi }{2}} - \left. {\left( {\sin x} \right)} \right|_{\frac{\pi }{2}}^\pi \)
\( = \left( {\sin \frac{\pi }{2} - \sin 0} \right) - \left( {\sin \pi - \sin \frac{\pi }{2}} \right) = 2\)