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How do you find the definite integral of #sqrt(y+1)*dy# from #[-1, 0]#?

1 Answer

Jul 5, 2016

#= 2/3#

Explanation:

#int_{-1)^0 dy qquad sqrt(y+1)#

this is simply the power rule ie #int du qquad u^n = n^{n+1}/(n+1)#

So

#=[2/3 (y+1)^(3/2)]_{-1)^0#

#=[2/3 (0+1)^(3/2)] - [2/3 (-1+1)^(3/2)] = 2/3#

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