Mister Exam

Other calculators

Solving integrals/ t*exp(-t/2)

Integral of t*exp(-t/2) dt

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
 oo          
  /          
 |           
 |     -t    
 |     ---   
 |      2    
 |  t*e    dt
 |           
/            
0
∫0∞te(−1)t2 dt\int\limits_{0}^{\infty} t e^{\frac{\left(-1\right) t}{2}}\, dt 
Integral(t*exp((-t)/2), (t, 0, oo))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(t)=tu{\left(t \right)} = t and let dv⁡(t)=e−t2\operatorname{dv}{\left(t \right)} = e^{- \frac{t}{2}}.

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

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

    1. Let u=−t2u = - \frac{t}{2}.

      Then let du=−dt2du = - \frac{dt}{2} and substitute −2du- 2 du:

      ∫(−2eu) du\int \left(- 2 e^{u}\right)\, du

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

        False\text{False}

        1. The integral of the exponential function is itself.

          ∫eu du=eu\int e^{u}\, du = e^{u}

        So, the result is: −2eu- 2 e^{u}

      Now substitute uu back in:

      −2e−t2- 2 e^{- \frac{t}{2}}

    Now evaluate the sub-integral.

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

    ∫(−2e−t2) dt=−2∫e−t2 dt\int \left(- 2 e^{- \frac{t}{2}}\right)\, dt = - 2 \int e^{- \frac{t}{2}}\, dt

    1. Let u=−t2u = - \frac{t}{2}.

      Then let du=−dt2du = - \frac{dt}{2} and substitute −2du- 2 du:

      ∫(−2eu) du\int \left(- 2 e^{u}\right)\, du

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

        False\text{False}

        1. The integral of the exponential function is itself.

          ∫eu du=eu\int e^{u}\, du = e^{u}

        So, the result is: −2eu- 2 e^{u}

      Now substitute uu back in:

      −2e−t2- 2 e^{- \frac{t}{2}}

    So, the result is: 4e−t24 e^{- \frac{t}{2}}

  3. Now simplify:

    −(2t+4)e−t2- \left(2 t + 4\right) e^{- \frac{t}{2}}

  4. Add the constant of integration:

    −(2t+4)e−t2+constant- \left(2 t + 4\right) e^{- \frac{t}{2}}+ \mathrm{constant}

The answer is:

−(2t+4)e−t2+constant- \left(2 t + 4\right) e^{- \frac{t}{2}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                                  
 |    -t              -t         -t 
 |    ---             ---        ---
 |     2               2          2 
 | t*e    dt = C - 4*e    - 2*t*e   
 |                                  
/
∫te(−1)t2 dt=C−2te−t2−4e−t2\int t e^{\frac{\left(-1\right) t}{2}}\, dt = C - 2 t e^{- \frac{t}{2}} - 4 e^{- \frac{t}{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
4
44 
=
= 
4
44 
4

    Use the examples entering the upper and lower limits of integration.

    © Mister Exam – Calculator