Question #674da

1 Answer
sente
Nov 25, 2016

#d/dxint_0^(2x)(e^t+2t)dt=2e^(2x)+8x#

Explanation:

#d/dxint_0^(2x)(e^t+2t)dt = d/dx[int_0^(2x)e^tdt+int_0^(2x)2tdt]#

#=d/dx[(e^t)_0^(2x)+(t^2)_0^(2x)]#

#=d/dx(e^(2x)-e^0+(2x)^2-0^2)#

#=d/dx(e^(2x)+4x^2-1)#

#=2e^(2x)+8x#

Note that we could have avoided calculating the integral and derivatives by letting #F(t) = int(e^t+2t)dt#, meaning #F'(t) = e^t+2t#. Then we have

#d/dxint_0^(2x)(e^t+2t)dt = d/dx[F(2x)-F(0)]#

#=2F'(2x)-0#

#=2[e^(2x)+2(2x)]#

#=2e^(2x)+8x#

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