Mister Exam

Other calculators

(x+2)*e^(-x)
Solving integrals/ (x+2)*e^(-x)

Integral of (x+2)*e^(-x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1               
  /               
 |                
 |           -x   
 |  (x + 2)*e   dx
 |                
/                 
0
∫01(x+2)e−x dx\int\limits_{0}^{1} \left(x + 2\right) e^{- x}\, dx 
Integral((x + 2)/E^x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    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(x)=x+2u{\left(x \right)} = x + 2 and let dv⁡(x)=e−x\operatorname{dv}{\left(x \right)} = e^{- x}.

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

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

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=−xu = - x.

          Then let du=−dxdu = - dx and substitute −du- du:

          ∫eu du\int e^{u}\, du

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

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

            1. The integral of the exponential function is itself.

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

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

          Now substitute uu back in:

          −e−x- e^{- x}

        Method #2

        1. Let u=e−xu = e^{- x}.

          Then let du=−e−xdxdu = - e^{- x} dx and substitute −du- du:

          ∫1 du\int 1\, du

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

            ∫(−1) du=−∫1 du\int \left(-1\right)\, du = - \int 1\, du

            1. The integral of a constant is the constant times the variable of integration:

              ∫1 du=u\int 1\, du = u

            So, the result is: −u- u

          Now substitute uu back in:

          −e−x- e^{- x}

      Now evaluate the sub-integral.

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

      ∫(−e−x) dx=−∫e−x dx\int \left(- e^{- x}\right)\, dx = - \int e^{- x}\, dx

      1. Let u=−xu = - x.

        Then let du=−dxdu = - dx and substitute −du- du:

        ∫eu du\int e^{u}\, du

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

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

          1. The integral of the exponential function is itself.

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

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

        Now substitute uu back in:

        −e−x- e^{- x}

      So, the result is: e−xe^{- x}

    Method #2

    1. Rewrite the integrand:

      (x+2)e−x=xe−x+2e−x\left(x + 2\right) e^{- x} = x e^{- x} + 2 e^{- x}

    2. Integrate term-by-term:

      1. Let u=−xu = - x.

        Then let du=−dxdu = - dx and substitute dudu:

        ∫ueu du\int u e^{u}\, du

        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(u)=uu{\left(u \right)} = u and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

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

          1. The integral of the exponential function is itself.

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

          Now evaluate the sub-integral.

        2. The integral of the exponential function is itself.

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

        Now substitute uu back in:

        −xe−x−e−x- x e^{- x} - e^{- x}

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

        ∫2e−x dx=2∫e−x dx\int 2 e^{- x}\, dx = 2 \int e^{- x}\, dx

        1. Let u=−xu = - x.

          Then let du=−dxdu = - dx and substitute −du- du:

          ∫eu du\int e^{u}\, du

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

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

            1. The integral of the exponential function is itself.

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

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

          Now substitute uu back in:

          −e−x- e^{- x}

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

      The result is: −xe−x−3e−x- x e^{- x} - 3 e^{- x}

  2. Now simplify:

    (−x−3)e−x\left(- x - 3\right) e^{- x}

  3. Add the constant of integration:

    (−x−3)e−x+constant\left(- x - 3\right) e^{- x}+ \mathrm{constant}

The answer is:

(−x−3)e−x+constant\left(- x - 3\right) e^{- x}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                      
 |                                       
 |          -x           -x            -x
 | (x + 2)*e   dx = C - e   - (2 + x)*e  
 |                                       
/
∫(x+2)e−x dx=C−(x+2)e−x−e−x\int \left(x + 2\right) e^{- x}\, dx = C - \left(x + 2\right) e^{- x} - e^{- x}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
       -1
3 - 4*e
3−4e3 - \frac{4}{e} 
=
= 
       -1
3 - 4*e
3−4e3 - \frac{4}{e}
Упростить
Numerical answer [src] 
1.52848223531423
1.52848223531423
The graph
Integral of (x+2)*e^(-x) dx

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

    © Mister Exam – Calculator