Cho hàm số \(f\left( x \right) = \sin 2x\) liên tục trên \(\mathbb{R}\).

14/22

a)\(\int\limits_0^\pi {f\left( x \right)dx} = 0\).

b) Biết \(F\left( 0 \right) = \frac{1}{2}\) thì \(F\left( {\frac{\pi }{2}} \right) = 1\).

c)\(\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x - f\left( x \right)} \right)dx = 2} \).

d)\(\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx = 4} \).

Nhập câu trả lời ở đây

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Giải thích

a) Đ, b) S, c) S, d) Đ

a) \(\int\limits_0^\pi {f\left( x \right)dx} \)\( = \int\limits_0^\pi {\sin 2xdx} \)\( = \left. {\left( { - \frac{{\cos 2x}}{2}} \right)} \right|_0^\pi \)\( = - \frac{1}{2} + \frac{1}{2} = 0\).

b) Ta có \(\int\limits_0^{\frac{\pi }{2}} {\sin 2x} dx = F\left( {\frac{\pi }{2}} \right) - F\left( 0 \right) = 1 \Rightarrow F\left( {\frac{\pi }{2}} \right) = 1 + F\left( 0 \right) = 1 + \frac{1}{2} = \frac{3}{2}\).

c) \(\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x - f\left( x \right)} \right)dx = } \)\[\int\limits_0^{\frac{\pi }{2}} {\cos xdx - \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)dx} } \]\[ = \int\limits_0^{\frac{\pi }{2}} {\cos xdx - \int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} } \]\[ = 1 - 1 = 0\].

d) \(\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx} \)\( = 2\int\limits_0^\pi {\left| {f\left( x \right)} \right|dx} \)\( = 2\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)dx} - 2\int\limits_{\frac{\pi }{2}}^\pi {f\left( x \right)dx} \)\( = 2\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} - 2\int\limits_{\frac{\pi }{2}}^\pi {\sin 2xdx} \)

\( = 2 + 2 = 4\).