Mathematics, 16.09.2019 16:30 alananicoleee
Since youtube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be approximated by v(t) = 525,600(0.42t2+2.7t_1) hours of video per year (0.5 _ t _ 7),
where t is time in years since the start of 2005.ê use a definite integral to estimate the total number of hours of video uploaded from the start of 2006 to the start of 2012. (round your answer to the nearest million hours of video.)
million hours of vie
Answers: 1
![\int\limits^7_1 {525,600(0.42t^2+2.7t-1)} \, dx \\=525,600 \left \{\frac{0.42t^3}{3}+\frac{2.7t^2}{2}-t \right. \left \} {{7} \atop {1}} \right. \\=525,600\{[0.14(343)+1.35(49)-7]-[0.14+1.35-1]\}\\=525,600(107.17-0.49)\\=525,600(106.68)\\=56,071,008](/tpl/images/0233/4783/88a43.png)
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Since youtube first became available to the public in mid-2005, the rate at which video has been upl...
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