Giải bài tập 3 trang 20 SGK Toán 12 tập 2 - Chân trời sáng tạo — Không quảng cáo

Đề bài

Tính các tích phân sau:

a) $\int\limits_{ - 2}^4 {\left( {x + 1} \right)\left( {x - 1} \right)dx}$

b) $\int\limits_1^2 {\frac{{{x^2} - 2x + 1}}{x}dx}$

c) $\int\limits_0^{\frac{\pi }{2}} {\left( {3\sin x - 2} \right)dx}$

d) $\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x}}{{1 + \cos x}}dx}$

Lời giải chi tiết

a) $\int\limits_{ - 2}^4 {\left( {x + 1} \right)\left( {x - 1} \right)dx}  = \int\limits_{ - 2}^4 {\left( {{x^2} - 1} \right)} dx = \left. {\left( {\frac{{{x^3}}}{3} - x} \right)} \right|_{ - 2}^4 = \left( {\frac{{{4^3}}}{3} - 4} \right) - \left( {\frac{{{{\left( { - 2} \right)}^3}}}{3} - \left( { - 2} \right)} \right) = 18$

b) $\int\limits_1^2 {\frac{{{x^2} - 2x + 1}}{x}dx}  = \int\limits_1^2 {\left( {x - 2 + \frac{1}{x}} \right)dx = \left. {\left( {\frac{{{x^2}}}{2} - 2x + \ln \left| x \right|} \right)} \right|_1^2}$

$= \left( {\frac{{2{\rm{^2}}}}{2} - 2.2 + \ln \left| 2 \right|} \right) - \left( {\frac{{1{\rm{^2}}}}{2} - 1.2 + \ln \left| 1 \right|} \right) = \ln 2 - \frac{1}{2}$

c) $\int\limits_0^{\frac{\pi }{2}} {\left( {3\sin x - 2} \right)dx}  = 3\int\limits_0^{\frac{\pi }{2}} {\sin xdx}  - 2\int\limits_0^{\frac{\pi }{2}} {dx}  = 3\left. {\left( { - \cos x} \right)} \right|_0^{\frac{\pi }{2}} - 2\left. {\left( x \right)} \right|_0^{\frac{\pi }{2}}$

$= 3\left[ {\left( { - \cos \frac{\pi }{2}} \right) - \left( { - \cos 0} \right)} \right] - 2\left( {\frac{\pi }{2} - 0} \right) = 3 - \pi$

d) $\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x}}{{1 + \cos x}}dx}  = \int\limits_0^{\frac{\pi }{2}} {\frac{{1 - {{\cos }^2}x}}{{1 + \cos x}}dx}  = \int\limits_0^{\frac{\pi }{2}} {\frac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{1 + \cos x}}dx = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos x} \right)dx} }$

$= \left. {\left( {x - \sin x} \right)} \right|_0^{\frac{\pi }{2}} = \left( {\frac{\pi }{2} - \sin \frac{\pi }{2}} \right) - \left( {0 - \sin 0} \right) = \frac{\pi }{2} - 1$