improper integrals

**saiyanmx89** · March 7th 2009, 04:48 AM

$\int e^{2x}dx$ [-infinity,0]

I set this equal to the limit as x goes to -infinity and subbed in a for -infinity. Then I solved the definite improper integral getting= $\frac{e^{2x}}{2}$ . I got (1/2)-0 = 1/2. Did I do this right??

**Mush** · March 7th 2009, 05:06 AM

Originally Posted by saiyanmx89

$\int e^{2x}dx$ [-infinity,0]

I set this equal to the limit as x goes to -infinity and subbed in a for -infinity. Then I solved the definite improper integral getting= $\frac{e^{2x}}{2}$ . I got (1/2)-0 = 1/2. Did I do this right??

Correct.

**skeeter** · March 7th 2009, 05:06 AM

Originally Posted by saiyanmx89

$\int e^{2x}dx$ [-infinity,0]

I set this equal to the limit as x goes to -infinity and subbed in a for -infinity. Then I solved the definite improper integral getting= $\frac{e^{2x}}{2}$ . I got (1/2)-0 = 1/2. Did I do this right??

$\lim_{a \to -\infty} \int_a^0 e^{2x} \, dx$

$\lim_{a \to -\infty} \left[\frac{1}{2}e^{2x}\right]_a^0$

$\lim_{a \to -\infty} \left[\frac{1}{2} - \frac{1}{2}e^{2a}\right] = \frac{1}{2}$

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