How do you evaluate the definite integral #int (2-t)sqrttdt# from [0,2]?

1 Answer
Jim G.
Mar 3, 2017

#16/15sqrt2#

Explanation:

#int_0^2(2-t)t^(1/2)dt#

#=int_0^2(2t^(1/2)-t^(3/2))dt#

integrate each term using the #color(blue)"power rule for integration"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(int(ax^n)=a/(n+1)x^(n+1);n≠-1)color(white)(2/2)|)))#

#rArrint_0^2(2t^(1/2)-t^(3/2))dt#

#=[4/3t^(3/2)-2/5t^(5/2)]_0^2#

#=(4/3 .2^(3/2)-2/5 .t^(5/2))-0#

#=4/3xx2sqrt2-2/5xx4sqrt2#

#=8/3sqrt2-8/5sqrt2#

#=16/15sqrt2#

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