How do you find the definite integral of #2x* (x-3)^5 dx# from #[-1, 2]#?

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Answer 1 · 2016-07-02T14:09:04

# = 4101/7#

we'll use IBP

#int u v' = uv - int u'v#

so for
#int_{-1}^2 dx qquad 2x (x-3)^5 #

we can set

#u = 2x, u' = 2#
#v' = (x-3)^5, v = 1/6(x-3)^6#

#implies [2x * 1/6(x-3)^6]\_(-1)^(2) - int_{-1}^2 dx qquad 2 * 1/6(x-3)^6 #

# =[1/3 x (x-3)^6]\_(-1)^(2) - 1/3 int_{-1}^2 dx qquad (x-3)^6#

# =[1/3 x (x-3)^6 - 1/3* 1/7(x-3)^7 ]\_(-1)^(2)#

# =[1/3 x (x-3)^6 - 1/21(x-3)^7 ]\_(-1)^(2)#

#= 2/3 + 1/21 + 4^6/3 - 4^7/21#

# = 4101/7#