Question

integrate (e^-x + e^-2x) dx / x from 0 to infinity​

Answer 1

Answer:

How do I integrate (x^2) × (e^-x) with limits 0 to infinite?

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SusaiRaj

Answered 1 year ago

We know that integ. udv= uv - integ.vdu

Let I = integ.x^2.e^(-x) dx

Let u = x^2 => du = 2x dx

dv = e^(-x) dx => v = -e^(-x)

Applying the above formula we get

I = -x^2.e^(-x) +integ.e^(-x) 2x dx

= -x^2.e^(-x) +2.I1 where I1 = integ.xe^(-x)dx..(1)

I1 = integ. x e(-x) dx

Let u = x => du = dx , dv = e^(-x)dx => v = -e^(-x)

Applying the above formula

I1 = x e^(-x) + integ.e^(-x) dx

= xe^(-x) - e^(-x) + c . Using this in (1) we get

I=-x^2.e^(-x) + 2{xe^(-x) - e^(-x)} + c

= x^2.e^(-x)+2x.e^(-x) - e^(-x) + c.

OR by using the formula

udv = uv - u'v1 + u''v2 - u'''v3….. we can directly write the answer by take ng u= x^2 and ..dv=e^(-x) dx. In the above u' stands for du/dx and v1 stands for integ .v dx

Answer 2

Step-by-step explanation:

First without the bounds:

I=∫e−2xdx

With u=−2x and du=−2dx:

I=−12∫e−2x(−2dx)=−12∫eudu=−12eu+C

Thus:

I=−12e−2x+C=−12e2x+C

Applying the bounds:

J=∫∞0e−2xdx=[−12e2x]∞0

Evaluating and taking the limit at infinity:

J=[