Integral of t*exp(-t) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - t$.

      Then let $du = - dt$ and substitute $du$:

      $$\int u e^{u}\, du$$

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

         Then $\operatorname{du}{\left(u \right)} = 1$.

         To find $v{\left(u \right)}$:

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         Now evaluate the sub-integral.

      2. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      Now substitute $u$ back in:

      $$- t e^{- t} - e^{- t}$$

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(t \right)} = t$ and let $\operatorname{dv}{\left(t \right)} = e^{- t}$.

      Then $\operatorname{du}{\left(t \right)} = 1$.

      To find $v{\left(t \right)}$:

      1. Let $u = - t$.

         Then let $du = - dt$ and substitute $- du$:

         $$\int \left(- e^{u}\right)\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\text{False}$$

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $- e^{u}$

         Now substitute $u$ back in:

         $$- e^{- t}$$

      Now evaluate the sub-integral.

   2. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- e^{- t}\right)\, dt = - \int e^{- t}\, dt$$

      1. Let $u = - t$.

         Then let $du = - dt$ and substitute $- du$:

         $$\int \left(- e^{u}\right)\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\text{False}$$

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $- e^{u}$

         Now substitute $u$ back in:

         $$- e^{- t}$$

      So, the result is: $e^{- t}$

2. Now simplify:

   $$- \left(t + 1\right) e^{- t}$$

3. Add the constant of integration:

   $$- \left(t + 1\right) e^{- t}+ \mathrm{constant}$$

The answer is:

$$- \left(t + 1\right) e^{- t}+ \mathrm{constant}$$