Помогите, пожалуйста. Created by kabzyapparova. algebra-ru

Помогите, пожалуйста...

Madina1908 $\int\limits^3_1 {2t^2+t} \, dt= \frac{2t^3}{3} + \frac{t^2}{2} \int\limits^3_1= \frac{2*3^3}{3} + \frac{3^2}{2} -( \frac{2}{3} + \frac{1}{2} )=18+ \frac{9}{2} - \frac{2}{3} -\frac{1}{2} =\\=18+4- \frac{2}{3} =21 \frac{1}{3}$
Ответ: $21 \frac{1}{3}$ м

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AnonimusPro

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S(х) = v ′(t)=2t - 2 S1= 2*1-2=0 S2= 2*3-2=4 S =S2-S1= 4-0=4

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