How do you evaluate the definite integral #int (x-sqrtx)/3dx# from [0,1]?

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Answer 1 · 2017-02-28T16:30:06

#-1/18.#

By the Fundamental Theorem of Calculus,

#int_0^1 (x-sqrtx)/3 dx=1/3[x^2/2-x^(1/2+1)/(1/2+1)]_0^1#

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

#=1/3{{1/2-2/3}-0]#

#=1/3(-1/6)#

#=-1/18.#

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