Question

Question #674da

Answer 1

`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`

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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`