How do you find the definite integral for: #(x+1)^(2)dx# for the intervals #[-7, 0]#?

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Answer 1 · 2016-06-08T01:32:56

#int_(-7)^0 (x+1)^2*dx=217/3#

The solution #1.# Use the power formula #int u^n*du=u^(n+1)/(n+1)#
#int_(-7)^0 (x+1)^2*dx#

Let #u=x+1# and #n=2# and #du=d(x+1)=dx#

#int_(-7)^0 (x+1)^2dx=[(x+1)^(2+1)/(2+1)]_(-7)^0#

#int_(-7)^0 (x+1)^2dx=[((x+1)^3)/3]_(-7)^0#

#int_(-7)^0 (x+1)^2dx=((0+1)^3)/3-((-7+1)^3)/3#

#int_(-7)^0 (x+1)^2dx=1/3-(-216)/3#

#color(red)(int_(-7)^0 (x+1)^2dx=217/3)#

The solution #2.# Expand the integrand

#int_(-7)^0 (x+1)^2dx=int_(-7)^0 (x^2+2x+1)*dx#

#int_(-7)^0 (x+1)^2dx=int_(-7)^0 x^2*dx+int_(-7)^0 2x dx+int_(-7)^0*dx#

#int_(-7)^0 (x+1)^2dx=[x^3/3+ x^2 +x]_(-7)^0#

#int_(-7)^0 (x+1)^2dx=[0^3/3+ 0^2 +0-((-7)^3/3+ (-7)^2 +(-7))]#

#int_(-7)^0 (x+1)^2dx=[0-(-343/3+ 49 -7)]#

#int_(-7)^0 (x+1)^2dx=+343/3- 49 +7#

#color(red)(int_(-7)^0 (x+1)^2dx=217/3)#

God bless....I hope the explanation is useful.