How do you solve #\int_{0}^{1} ( 1+ \sqrt{x} )^8 dx#?

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Answer 1 · 2016-10-01T22:36:36

#4097/45#

Making #y = sqrt(x)# in #\int_{0}^{1} ( 1+ \sqrt{x} )^8 dx# and considering #dy = 1/2 dx/sqrt(x) = 1/2 dx/y# we have

#\int_{0}^{1} ( 1+ \sqrt{x} )^8 dx equiv 2 int_0^1(1+y)^8y dy#

but #(1+y)^9 = (1+8)^8(1+y) = (1+y)^8+(1+y)^8 y#

so

#2 int_0^1(1+y)^8y dy = 2int_0^1(1+y)^9dy-2int_0^1(1+y)^8dy#

so

#2 int_0^1(1+y)^8y dy = (2/10(1+y)^10 )_0^1 -(2/9(1+y)^9)_0^1=4097/45#