Lời giải
\( v(t) = 0 \Rightarrow t = \frac{1}{2} \) (s)
\( S(t) = \int v(t) dt = \int (-40t + 20) dt = -20t^2 + 20t + C \)
\( S(0) = 0 \Rightarrow C = 0 \)
\( S(t) = -20t^2 + 20t \Rightarrow S( \frac{1}{2}) = 10 - 5 = 5 \) m
\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{D}} \)
page 59
(2017. Câu 35)
Lời giải
* \( v(t) = at^2 + bt \)
\( \begin{cases} 4a + 2b = 9 \\ -\frac{b}{2a} = 2 \end{cases}
\Leftrightarrow \begin{cases} b = -4a\\ a = -\frac{9}{4} \end{cases}
\Leftrightarrow \begin{cases} a = -\frac{9}{4}\\ b = 9 \end{cases} \)
\( v(t) = -\frac{9}{4}t^2 + 9t \)
\( t = 3 \Rightarrow v = \frac{27}{4} \)
\( \Rightarrow S(t) = \int_0^3 (-\frac{9}{4}t^2 + 9t) dt + \int_3^4 \frac{27}{4} dt = 27 \)
\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{C}} \)
page 60
\( x^2 + (y - 2)^2 = 1 \Leftrightarrow (y = 2)^2 = 1 - x^2 \)
\( \Leftrightarrow y = 2 \pm \sqrt{1 - x^2} \)
\( V = \pi \int_{-1}^{1} [(2 + \sqrt{1 - x^2})^2 - (2 - \sqrt{1 - x^2})^2] dx \)
\( = \pi \int_{-1}^{1} 8\sqrt{1 - x^2} \, dx \)
Đặt x = Sin t
\( V = 4\pi^2 \)(dvtt)
page 61
* \( x^2 = \frac{27}{x} \Rightarrow x = 3 \)
* \( \frac{x^2}{27} = \frac{27}{x} \Rightarrow x = 9 \)
a) \( V = \pi \int_0^3 \left( x^4 - \frac{x^4}{27^2} \right) dx + \pi \int_3^9 \left[ \left(\frac{27}{x}\right)^2 - \left(\frac{x^2}{27}\right)^2 \right] dx\)
b) \(V = \pi \int_0^3 \left( \left(\sqrt{27y}\right)^2 - (\sqrt{y})^2 \right) dy + \pi \int_3^9 \left[ \left(\frac{27}{y}\right)^2 - (\sqrt{y})^2 \right] dy\)
page 62
\( y^2 = x^3 \Leftrightarrow y = \pm \sqrt{x^3} \Leftrightarrow x = \sqrt[3]{y^2} \)
(a) \( V = \pi \int_0^1 x^3 \, dx \)
(b) \( V = \pi \int_0^1 \left( 1 - y^{\frac{4}{3}} \right) dy \)
page 63