Tích phân $\Large I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{4

Tích phân $\Large I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{4x-3}{\sqrt{5+4x-{{x}^{2}}}}dx}$ có giá trị là :

• A.

  $\Large I=\dfrac{5\pi }{3}$

• B.

  $\Large I=\dfrac{5\pi }{6}$

• C.

  $\Large I=-\dfrac{5\pi }{3}$

• D.

  $\Large I=-\dfrac{5\pi }{6}$

Ta có: $\Large (5+4x-{{x}^{2}}{)}'=4-2x$ và $\Large 4x-3=5-2(4-2x)$

$\Large I=\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{4x-3}{\sqrt{5+4x-{{x}^{2}}}}dx=\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{5}{\sqrt{5+4x-{{x}^{2}}}}dx-\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{2(4-2x)}{\sqrt{5+4x-{{x}^{2}}}}dx}}}$

Xét $\Large {{I}_{1}}=\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{5}{\sqrt{5+4x-{{x}^{2}}}}dx=\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{5}{\sqrt{9-{{(x-2)}^{2}}}}dx}}$

Đặt $\Large x-2=3\sin t,t\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right]\Rightarrow dx=3\cos tdt$

Đổi cận: $\Large \left\{ \begin{align}  & x=\dfrac{7}{2}\Rightarrow t=\frac{\pi }{6} \\  & x=\dfrac{1}{2}\Rightarrow t=-\dfrac{\pi }{6} \\ \end{align} \right.$

$\Large \Rightarrow {{I}_{1}}=\int\limits_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\dfrac{5.3\cos t}{\sqrt{9-9{{\sin }^{2}}t}}dt=\dfrac{5\pi }{3}}$

Xét $\Large {{I}_{2}}=\int\limits_{\frac{1}{2}}^{\frac{7}{2}}{\dfrac{2(4-2x)}{\sqrt{5+4x-{{x}^{2}}}}dx}$

Đặt $\Large t=5+4x-{{x}^{2}}\Rightarrow dt=4-2x$

Đổi cận: $\Large \left\{ \begin{align}  & x=\dfrac{1}{2}\Rightarrow t=\dfrac{27}{4} \\  & x=\dfrac{7}{2}\Rightarrow t=\dfrac{27}{4} \\ \end{align} \right.$ $\Large \Rightarrow {{I}_{2}}=0$

$\Large \Rightarrow I=\dfrac{5\pi }{3}$

Chọn A