Integral of (x^5)e^(-x^2) dx
The solution
Detail solution
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Let u=−x2.
Then let du=−2xdx and substitute −2du:
∫(−2u2eu)du
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The integral of a constant times a function is the constant times the integral of the function:
∫u2eudu=−2∫u2eudu
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u2 and let dv(u)=eu.
Then du(u)=2u.
To find v(u):
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The integral of the exponential function is itself.
∫eudu=eu
Now evaluate the sub-integral.
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=2u and let dv(u)=eu.
Then du(u)=2.
To find v(u):
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The integral of the exponential function is itself.
∫eudu=eu
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫2eudu=2∫eudu
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The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 2eu
So, the result is: −2u2eu+ueu−eu
Now substitute u back in:
−2x4e−x2−x2e−x2−e−x2
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Now simplify:
−(2x4+x2+1)e−x2
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Add the constant of integration:
−(2x4+x2+1)e−x2+constant
The answer is:
−(2x4+x2+1)e−x2+constant
The answer (Indefinite)
[src]
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| 2
| 2 2 2 4 -x
| 5 -x -x 2 -x x *e
| x *E dx = C - e - x *e - -------
| 2
/
∫e−x2x5dx=C−2x4e−x2−x2e−x2−e−x2
The graph
1−2e5
=
1−2e5
Use the examples entering the upper and lower limits of integration.