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Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{ll}x & \text { khi } x \geq \dfrac{\pi}{2} \\ \cos x & \text { khi } x<\dfrac{\pi}{2}\end{array}\right.$. Biết tích phân $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=\dfrac{\pi}{a}+b$ (với $a, b \in$ Z, $a \neq 0$ ). Tính $S=a+b$.
A. $S=5$.
B. $S=-3$.
C. $S=3$.
D. $S=-5$.
A. $S=5$.
B. $S=-3$.
C. $S=3$.
D. $S=-5$.
Ta có $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=\int_0^{\dfrac{\pi}{2}} \cos ^2 x \mathrm{~d} x+\int_{\dfrac{\pi}{2}}^\pi x \cdot \cos x \mathrm{~d} x$.
+) Tính $A=\int_0^{\dfrac{\pi}{2}} \cos ^2 x \mathrm{~d} x=\dfrac{1}{2} \int_0^{\dfrac{\pi}{2}}(1+\cos 2 x) \mathrm{d} x=\left.\dfrac{1}{2}\left(x+\dfrac{1}{2} \sin 2 x\right)\right|_0 ^{\dfrac{\pi}{2}}=\dfrac{\pi}{4}$.
+) Tính $B=\int_{\dfrac{\pi}{2}}^\pi x \cdot \cos x \mathrm{~d} x=\int_{\dfrac{\pi}{2}}^\pi x \mathrm{~d}(\sin x)=\left.x \cdot \sin x\right|_{\dfrac{\pi}{2}} ^\pi-\int_{\dfrac{\pi}{2}}^\pi \sin x \mathrm{~d} x$
$=\left(\pi \cdot \sin \pi-\dfrac{\pi}{2} \sin \dfrac{\pi}{2}\right)+\left.\cos x\right|_{\dfrac{\pi}{2}} ^\pi=-\dfrac{\pi}{2}+\left(\cos \pi-\cos \dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}-1$.
Suy ra $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=A+B=\dfrac{\pi}{4}+\left(-\dfrac{\pi}{2}-1\right)=-\dfrac{\pi}{4}-1$.
Mặc khác $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=\dfrac{\pi}{a}+b$. Ta có $a=-4, b=-1$.
Vậy $S=a+b=-5$.
+) Tính $A=\int_0^{\dfrac{\pi}{2}} \cos ^2 x \mathrm{~d} x=\dfrac{1}{2} \int_0^{\dfrac{\pi}{2}}(1+\cos 2 x) \mathrm{d} x=\left.\dfrac{1}{2}\left(x+\dfrac{1}{2} \sin 2 x\right)\right|_0 ^{\dfrac{\pi}{2}}=\dfrac{\pi}{4}$.
+) Tính $B=\int_{\dfrac{\pi}{2}}^\pi x \cdot \cos x \mathrm{~d} x=\int_{\dfrac{\pi}{2}}^\pi x \mathrm{~d}(\sin x)=\left.x \cdot \sin x\right|_{\dfrac{\pi}{2}} ^\pi-\int_{\dfrac{\pi}{2}}^\pi \sin x \mathrm{~d} x$
$=\left(\pi \cdot \sin \pi-\dfrac{\pi}{2} \sin \dfrac{\pi}{2}\right)+\left.\cos x\right|_{\dfrac{\pi}{2}} ^\pi=-\dfrac{\pi}{2}+\left(\cos \pi-\cos \dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}-1$.
Suy ra $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=A+B=\dfrac{\pi}{4}+\left(-\dfrac{\pi}{2}-1\right)=-\dfrac{\pi}{4}-1$.
Mặc khác $I=\int_0^\pi f(x) \cdot \cos x \mathrm{~d} x=\dfrac{\pi}{a}+b$. Ta có $a=-4, b=-1$.
Vậy $S=a+b=-5$.
Đáp án D.