Giải bài 29 trang 53 sách bài tập toán 12 - Kết nối tri thức

Đề bài

a) \(\int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}\frac{x}{2}dx} \);

b) \(\int\limits_0^1 {\left( {3x - 4{x^3}} \right)dx}  + \int\limits_1^2 {\left( {4{x^3} - 3x} \right)dx} \);

c) \(\int\limits_0^6 {\left( {\left| {2x - 2} \right| + 4{x^2}} \right)dx} \).

Lời giải chi tiết

a) Ta có \(\int\limits_0^{\frac{\pi }{4}} {{{\sin }^2}\frac{x}{2}dx}  = \int\limits_0^{\frac{\pi }{4}} {\frac{{1 - \cos x}}{2}dx}  = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {dx}  - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos xdx}  = \left. {\frac{x}{2}} \right|_0^{\frac{\pi }{4}} - \frac{1}{2}\left. {\sin x} \right|_0^{\frac{\pi }{4}}\)\( = \frac{\pi }{8} - \frac{{\sqrt 2 }}{4}\).

b) Ta có

\(\int\limits_0^1 {\left( {3x - 4{x^3}} \right)dx}  + \int\limits_1^2 {\left( {4{x^3} - 3x} \right)dx}  = \left. {\left( {\frac{{3{x^2}}}{2} - {x^4}} \right)} \right|_0^1 + \left. {\left( {{x^4} - \frac{{3{x^2}}}{2}} \right)} \right|_1^2 = 11\).

c) Xét \(\left| {2x - 2} \right|\) trên \(\left[ {0;6} \right]\) có \(\left| {2x - 2} \right| = \left\{ \begin{array}{l}2x - 2,1 \le x \le 6\\ - 2x + 2,0 \le x < 1\end{array} \right.\).

Suy ra \(\int\limits_0^6 {\left( {\left| {2x - 2} \right| + 4{x^2}} \right)dx}  = \int\limits_0^1 {\left( { - 2x + 2 + 4{x^2}} \right)dx + } \int\limits_1^6 {\left( {2x - 2 + 4{x^2}} \right)dx} \)

\(\begin{array}{l} = \left. {\left( { - {x^2} + 2x + \frac{{4{x^3}}}{3}} \right)} \right|_0^1 + \left. {\left( {{x^2} - 2x + \frac{{4{x^3}}}{3}} \right)} \right|_1^6 =  - 1 + 2 + \frac{4}{3} + 36 - 12 + 288 - \left( {1 - 2 + \frac{4}{3}} \right)\\ = \frac{{943}}{3} - \frac{1}{3} = 314.\end{array}\)