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Câu hỏi: Biết $\int\limits_{0}^{2}{\left( 3x-1 \right)}{{e}^{\dfrac{x}{2}}}dx=a+be$, với $a,b$ là số hữa tỉ. Tính ${{a}^{2}}-{{b}^{2}}$.
A. $192.$
B. $-192.$
C. $200.$
D. $-200.$
A. $192.$
B. $-192.$
C. $200.$
D. $-200.$
Xét $\int\limits_{0}^{2}{\left( 3x-1 \right){{\text{e}}^{\dfrac{x}{2}}}\text{d}x}$.
Đặt $u=\left( 3x-1 \right)\Rightarrow \text{d}u=3\text{d}x$ ; $\text{d}v={{\text{e}}^{\dfrac{x}{2}}}\text{d}x\Rightarrow v=2{{\text{e}}^{\dfrac{x}{2}}}$.
Vậy ta có: $\int\limits_{0}^{2}{\left( 3x-1 \right){{\text{e}}^{\dfrac{x}{2}}}\text{d}x}=\left. 2{{\text{e}}^{\dfrac{x}{2}}}\left( 3x-1 \right) \right|_{0}^{2}-6\int\limits_{0}^{2}{{{\text{e}}^{\dfrac{x}{2}}}\text{d}x}=\left. 2{{\text{e}}^{\dfrac{x}{2}}}\left( 3x-1 \right) \right|_{0}^{2}-\left. 12{{e}^{\dfrac{x}{2}}} \right|_{0}^{2}$ $=2\left( e\left( 3.2-1 \right)-{{e}^{0}}\left( 3.0-1 \right) \right)-12\left( e-{{e}^{0}} \right)$ $=10e+2-12e+12=14-2e$.
Vậy ta có $a=14;b=-2\Rightarrow {{a}^{2}}-{{b}^{2}}=192$.
Đặt $u=\left( 3x-1 \right)\Rightarrow \text{d}u=3\text{d}x$ ; $\text{d}v={{\text{e}}^{\dfrac{x}{2}}}\text{d}x\Rightarrow v=2{{\text{e}}^{\dfrac{x}{2}}}$.
Vậy ta có: $\int\limits_{0}^{2}{\left( 3x-1 \right){{\text{e}}^{\dfrac{x}{2}}}\text{d}x}=\left. 2{{\text{e}}^{\dfrac{x}{2}}}\left( 3x-1 \right) \right|_{0}^{2}-6\int\limits_{0}^{2}{{{\text{e}}^{\dfrac{x}{2}}}\text{d}x}=\left. 2{{\text{e}}^{\dfrac{x}{2}}}\left( 3x-1 \right) \right|_{0}^{2}-\left. 12{{e}^{\dfrac{x}{2}}} \right|_{0}^{2}$ $=2\left( e\left( 3.2-1 \right)-{{e}^{0}}\left( 3.0-1 \right) \right)-12\left( e-{{e}^{0}} \right)$ $=10e+2-12e+12=14-2e$.
Vậy ta có $a=14;b=-2\Rightarrow {{a}^{2}}-{{b}^{2}}=192$.
Đáp án A.
