a) \( V = \pi \int_0^4 (2^2 -y^2) \,dx= \pi \int_0^4 (4 - x) \, dx = \pi  \Big(4x - \frac{x^2}{2}\Big) \bigg|_0^4\, dx = 8\pi \)

b) \( V = \pi \int_0^2 x^2 \, dy = \pi \int_0^2 y^4 \, dy = \pi \frac{y^5}{5}\Big|_0^2 = \frac{32\pi}{5} \)

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\( -x^2 + 2x = 0  \Leftrightarrow \left[ \begin{array}{l} x = 0 \\ x = 2 \end{array} \right. \)

a) \( V = \pi \int_0^2 (-x^2 + 2x)^2 \, dx = \frac{16\pi}{15} \, \) (đvtt)

b) \( y = -x^2 + 2x = 1 - (x - 1)^2 \iff (x - 1)^2 = 1 - y \)

\( \Leftrightarrow x - 1 = \pm \sqrt{1 - y} \Leftrightarrow \left[ \begin{array}{l} x = 1 + \sqrt{1 - y} \\ x = 1 - \sqrt{1 - y} \end{array} \right. \)

\( V = \pi \int_0^1 \Big[ (1 + \sqrt{1 - y})^2 - (1 - \sqrt{1 - y})^2 \Big] \, dy \)

\( = \pi \int_0^1 \Big( 4\sqrt{1 - y} \Big) \, dy = 4\pi  (1 - y)^{\frac{3}{2}}(\frac{-2}{3}) \bigg|_0^1 = \frac{8\pi}{3}\) (đvtt)

\( V = \frac{128\pi}{3} \)

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Lời giải

•  \( x^2 - 2x + 2 = x^2 + 4x + 5 \Leftrightarrow x = -\frac{1}{2} \)

\( S = S_1 + S_2 = \int_{-2}^{\frac{-1}{2}} \big( x^2 + 4x + 5 - 1 \big) \, dx + \int_{\frac{-1}{2}}^1 \big( x^2 - 2x + 2 - 1 \big) \, dx \)

    \( = \int_{-2}^{\frac{-1}{2}} \big( x^2 + 4x + 4 \big) \, dx + \int_{\frac{-1}{2}}^1 \big( x^2 - 2x + 1 \big) \, dx \)

    \( = \left( \frac{x^3}{3} + 2x^2 + 4x \right) \bigg|_{-2}^{\frac{-1}{2}} + \left( \frac{x^3}{3} - x^2 + x \right) \bigg|_{\frac{-1}{2}}^1 \)

    \( = \frac{9}{4} \) (đvdt)

Bấm 

\( \int_{-2}^{\frac{-1}{2}} \big( x^2 + 4x + 1 \big) \, dx + \int_{\frac{-1}{2}}^1 \big( x^2 - 2x + 1 \big) \, dx = \frac{9}{4}\)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{C}} \)

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Lời giải

*  \( x^2 = \frac{8}{x} \iff x^3 = 8 \iff x = 2 \)

*  \( \frac{x^2}{8} = \frac{8}{x} \iff x^3 = 64 \iff x = 4 \)

\( S = S_1 + S_2 = \int_0^2 \big( x^2 - \frac{x^2}{8} \big) \, dx + \int_2^4 \big( \frac{8}{x} - \frac{x^2}{8} \big) \, dx \)

   \(  = 8\ln2  \)

Bấm
•  \( \int_0^2 \frac{7x^2}{8} \, dx + \int_2^4 \big( \frac{8}{x} - \frac{x^2}{8} \big) \, dx = 5.545177 \)

•  \( 8\ln2 = 5.545177\)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{C}} \)

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Lời giải

* \( e^{x-2} = 3 - x \Leftrightarrow x = 2 \) (đơn điệu)  

\( S = S_1 + S_2 = \int_0^2 e^{x-2} \, dx + \int_2^3 (3 - x) \, dx \)

\(= e^{x-2} \bigg|_0^2 + ( 3x - \frac{x^2}{2}) \bigg|_2^3  \)

\( = \left( 1 - \frac{1}{e^2} \right) + \left( \frac{9}{2} - 4 \right) = \frac{3}{2} - \frac{1}{e^2} \) (đvdt)

Bấm
•  \( \int_0^2 e^{x-2} \, dx + \int_2^3 (3 - x) \, dx = 1.364664\)

•  \( \frac{3}{2} - \frac{1}{e^2} = 1.364664 \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)

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