Integral of exponential function

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$\int e^{17-20x} \; dx$

**Answer**

$\int e^{17-20x} \; dx$

Substitute $u= 17-20x$. So, $du = -20\; dx$, or $dx =-\frac{1}{20}\; du $

Then $\int e^{17-20x} \; dx$

$=-\frac{1}{20}\int e^u \; du $

$=-\frac{1}{20}e^u + C $

Substituting back $u=17-20x$

$=-\frac{1}{20} e^{17-20x} + C$