   $$ \int_0^{\infty}\!ce^{-2c}\,\mathrm{d}x $$=1

It seems there is a small inconsistency in the integral you provided. Specifically, the variable of integration is written as $ \mathrm{d}x $, but the integrand uses the variable $ c $.

Let me assume you meant to integrate with respect to $ c $, since the integrand involves $ c $. If that’s correct, the integral would be:

\[
\int_0^{\infty} ce^{-2c} \, \mathrm{d}c
\]

### Solution:
We can solve this using integration by parts or using a standard integral form.

Let’s use integration by parts. Set:
- $ u = c $, so $ \mathrm{d}u = \mathrm{d}c $
- $ \mathrm{d}v = e^{-2c} \mathrm{d}c $, so integrating gives $ v = -\frac{1}{2}e^{-2c} $

Now, apply the integration by parts formula:
\[
\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u
\]

This gives:
\[
\int_0^{\infty} ce^{-2c} \, \mathrm{d}c = \left[-\frac{1}{2} c e^{-2c} \right]_0^{\infty} + \frac{1}{2} \int_0^{\infty} e^{-2c} \, \mathrm{d}c
\]

Evaluating the boundary terms:

At $ c = \infty $:
\[
\lim_{c \to \infty} -\frac{1}{2} c e^{-2c} = 0
\]

At $ c = 0 $:
\[
-\frac{1}{2} \cdot 0 \cdot e^{0} = 0
\]

Now for the remaining integral:
\[
\frac{1}{2} \int_0^{\infty} e^{-2c} \, \mathrm{d}c = \frac{1}{2} \left[ \frac{1}{2} \right] = \frac{1}{4}
\]

Thus, the value of the original integral is:
\[
\int_0^{\infty} ce^{-2c} \, \mathrm{d}c = \frac{1}{4}
\]

It looks like you might be referring to a different integral if you expected the value to be 1. Could you confirm if this is the correct integral, or if there’s an adjustment needed?

Would you like more details on integration by parts or a related concept?

### Follow-up Questions:
1. Would you like to see an alternative method of solving this integral?
2. Do you need a more detailed explanation of integration by parts?
3. Is there a specific application you are studying that involves integrals like this?
4. Should we explore how to use a substitution method for such integrals?
5. Do you need help generalizing this integral for other exponentials?

**Tip:** For integrals involving exponentials and powers of variables, integration by parts is a useful method to remember!

$$ \int_0^{\infty} ce^{-2c} \, \mathrm{d}x = 1 $$

This page provides a step-by-step solution to the integral $ \int_0^{\infty} ce^{-2c} \, \mathrm{d}c $ using the integration by parts method. Learn about key calculus concepts, such as integration by parts and exponential functions, with this detailed solution suitable for calculus students.

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