What is the integral of #e^(-0.2t) dt#?

Question

HIX Tutor · Accepted Answer

undefined To find the integral of $ e^{-0.2t} \, dt $, follow these steps:

1. **Recognize the formula**: The integral of $ e^{at} \, dt $ is $ \frac{1}{a} e^{at} + C $, where $ a $ is a constant and $ C $ is the constant of integration.

2. **Identify $ a $**: In this case, $ a = -0.2 $.

3. **Apply the formula**:
   - Use $ a = -0.2 $:
   $$
   \int e^{-0.2t} \, dt = \frac{1}{-0.2} e^{-0.2t} + C
   $$

4. **Simplify**:
   - Calculate $ \frac{1}{-0.2} = -5 $:
   $$
   \int e^{-0.2t} \, dt = -5 e^{-0.2t} + C
   $$

5. **Final answer**:
   $$
   \int e^{-0.2t} \, dt = -5 e^{-0.2t} + C
   $$

That's it!

Cameron Alexander · Answer

#int e^(-0.2t) dt = 5e^(-0.2t) +C#

Two methods:

Substitution

Let #u = 0.2t# so #du = -0.2 dt#

#int e^(-0.2t) dt# becomes

#-1/0.2 int e^u du = -1/(2/10) e^u +C#

# = -10/2 e^(-0.2t) +C#

# = -5 e^(-0.2t) +C#

Check the answer by differentiating.

Learn the rule

#int e^(ku) du = 1/k e^(ku) +C#

(Verify be differentiating.)

So

#int e^(-0.2t) dt = 1/(-0.2)e^(-0.2t) +C#

# = -5 e^(-0.2t) +C#

Samantha Adkison · Answer

undefined The integral of $ e^{-0.2t} $ with respect to $ t $ is:

$$ \int e^{-0.2t} dt = -\frac{1}{0.2} e^{-0.2t} + C $$

Where $ C $ is the constant of integration. Therefore, the integral is:

$$ -\frac{1}{0.2} e^{-0.2t} + C $$