Question #674da

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Question #674da

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Answer 1 · 2016-11-25T09:59:40

$\frac{d}{d x} \int_{0}^{2 x} (e^{t} + 2 t) d t = 2 e^{2 x} + 8 x$ $d/dxint_0^(2x)(e^t+2t)dt=2e^(2x)+8x$

Explanation:

$\frac{d}{d x} \int_{0}^{2 x} (e^{t} + 2 t) d t = \frac{d}{d x} [\int_{0}^{2 x} e^{t} d t + \int_{0}^{2 x} 2 t d t]$ $d/dxint_0^(2x)(e^t+2t)dt = d/dx[int_0^(2x)e^tdt+int_0^(2x)2tdt]$

$= \frac{d}{d x} [{(e^{t})}_{0}^{t} + {(t^{2})}_{0}^{2}]$ $=d/dx[(e^t)_0^(2x)+(t^2)_0^(2x)]$

$= \frac{d}{d x} (e^{2 x} - e^{0} + {(2 x)}^{2} - 0^{2})$ $=d/dx(e^(2x)-e^0+(2x)^2-0^2)$

$= \frac{d}{d x} (e^{2 x} + 4 x^{2} - 1)$ $=d/dx(e^(2x)+4x^2-1)$

$= 2 e^{2 x} + 8 x$ $=2e^(2x)+8x$

Note that we could have avoided calculating the integral and derivatives by letting $F (t) = \int (e^{t} + 2 t) d t$ $F(t) = int(e^t+2t)dt$ , 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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Question #674da

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