What is #f(x) = int xsqrt(x-2) dx# if #f(1)=-2 #?

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Answer 1 · 2017-05-10T18:20:32

# f(x)=2/15(x-2)^(3/2)(3x+4)+C.#

#f(x)=intxsqrt(x-2)dx.#

We subst. #x-2=t^2, i.e., x=2+t^2 rArr dx=2tdt.#

#rArr I=int(t^2+2)sqrt(t^2)*2tdt#

#=2intt^2(t^2+2)dt=2int(t^4+2t^2)dt#

#=2[t^5/5+2*t^3/3]#

#=2/15*(3t^5+10t^3)#

#=2/15*t^3(3t^2+10)," and, since, "t=sqrt(x-2)," we have,"#

# f(x)=2/15*(x-2)^(3/2){3(x-2)+10}+C, or, #

# f(x)=2/15(x-2)^(3/2)(3x+4)+C.#