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Câu hỏi: Cho tích phân $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 4x-1+\cos x \right)\text{d}x}=\pi \left( \dfrac{\pi }{a}-\dfrac{1}{b} \right)+C$, $\left( a,b,C\in \mathbb{Q} \right)$. Tính $a-b+C$
A. $\dfrac{1}{2}$.
B. $1$.
C. $-2$.
D. $\dfrac{1}{3}$.
A. $\dfrac{1}{2}$.
B. $1$.
C. $-2$.
D. $\dfrac{1}{3}$.
Ta có $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 4x-1+\cos x \right)\text{d}x}=\left. \left( 2{{x}^{2}}-x+\sin x \right) \right|_{0}^{\dfrac{\pi }{2}}=\pi \left( \dfrac{\pi }{2}-\dfrac{1}{2} \right)+1$.
Suy ra $a=2$, $b=2$, $c=1$ nên $a-b+c=1$.
Suy ra $a=2$, $b=2$, $c=1$ nên $a-b+c=1$.
Đáp án B.