Cho hàm số f ( x ) có f ( pi/ 2 ) = 4 và f ′ ( x ) = 2 sin 2 x + 1 , ∀ x ∈ ( 0 ; π ) . Khi đó 3 pi/ 4 ∫ pi/ 4 f ( x ) d x bằng

4/22

Cho hàm số \[f(x)\]có \[f\left( {\frac{\pi }{2}} \right) = 4\]và\(f'\left( x \right) = \frac{2}{{{{\sin }^2}x}} + 1,\forall x \in \left( {0;\pi } \right)\). Khi đó \(\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {f\left( x \right){\rm{d}}x} \)bằng

\(\frac{{{\pi ^2}}}{2}\).

\(2\pi \).

\(\frac{{8\pi - {\pi ^2}}}{2}\).

\(\frac{{{\pi ^2} + 2\pi }}{2}\).

Giải thích

Đáp án đúng là: B

Ta có: \[\int {f'\left( x \right){\rm{d}}x} = \int {\left( {\frac{2}{{{{\sin }^2}x}} + 1} \right)} dx = - 2\cot x + x + C = f(x)\].

\[f\left( {\frac{\pi }{2}} \right) = 4 \Leftrightarrow C = 4 - \frac{\pi }{2} \Rightarrow f\left( x \right) = - 2\cot x + x + 4 - \frac{\pi }{2}\].

\(\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {f\left( x \right){\rm{d}}x} = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( { - 2\cot x + x + 4 - \frac{\pi }{2}} \right){\rm{d}}x} = = \left. {\left[ { - 2\ln \left( {\sin x} \right) + \frac{{{x^2}}}{2} + 4x - \frac{\pi }{2}x} \right]} \right|_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} = 2\pi \).