Question

Evaluate \(\rm \int \frac{1}{e^{2x}}dx\)
1. \(\rm \frac{-1}{2} {e^{2x}}+c\)
2. \(\rm \frac{1}{2} {e^{-2x}}+c\)
3. \(\rm \frac{-1}{2} {e^{-2x}}+c\)
4. None of the above

Answer 1

Correct Answer - Option 3 : \(\rm \frac{-1}{2} {e^{-2x}}+c\)

Concept:

\(\rm \int e^x dx = e^x + c\)

Calculation:

I = \(\rm \int \frac{1}{e^{2x}}dx\)

= \(\rm \int {e^{-2x}}dx\)

Let -2x = t

Differentiating with respect to x, we get

⇒ -2dx = dt

⇒ dx = \(\frac {-1}{2}\)dt

Now, 

I = \(\rm \frac{-1}{2}\int {e^{t}}dt\)

= \(\rm \frac{-1}{2} {e^{t}}+c\)

= \(\rm \frac{-1}{2} {e^{-2x}}+c\)