Hướng dẫn giải Câu hỏi Thực hành 5 trang 19 SGK Toán 12 Chân trời sáng tạo – Bài 2. Tích phân. Hướng dẫn: 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} \.
Câu hỏi/Đề bài:
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) \(\int\limits_0^\pi {\left| {\cos x} \right|dx} \)
Hướng dẫn:
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 \(\left| {\cos x} \right| = \left\{ {\begin{array}{*{20}{c}}{\cos x{\rm{ }}\left( {0 \le x \le \frac{\pi }{2}} \right)}\\{ – \cos x{\rm{ }}\left( {\frac{\pi }{2} \le x \le \frac{\pi }{2}} \right)}\end{array}} \right.\).
Từ đó ta có \(\int\limits_0^\pi {\left| {\cos x} \right|dx} = \int\limits_0^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} + \int\limits_{\frac{\pi }{2}}^\pi {\left| {\cos x} \right|dx} \).
Lời giải:
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) \(\int\limits_0^\pi {\left| {\cos x} \right|dx} = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} + \int\limits_{\frac{\pi }{2}}^\pi {\left( { – \cos x} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {\cos xdx} – \int\limits_{\frac{\pi }{2}}^\pi {\cos xdx} = \left. {\left( {\sin x} \right)} \right|_0^{\frac{\pi }{2}} – \left. {\left( {\sin x} \right)} \right|_{\frac{\pi }{2}}^\pi \)
\( = \left( {\sin \frac{\pi }{2} – \sin 0} \right) – \left( {\sin \pi – \sin \frac{\pi }{2}} \right) = 2\)