Question

Question #1f484

Answer 1

`intsqrtxe^(-sqrtx)dx=-2 e^(-sqrt(x)) (x + 2 sqrt(x) + 2) + c`

Explanation:

For `intsqrtxe^(-sqrtx)dx`, let `u=sqrtx` and `du=1/2sqrtxdx`

Then `intsqrtxe^(-sqrtx)dx=2intu^2e^udu`

For the integrand `u^2e^(-u)`, integrate by parts, `int f dg = f g - int g df`, where
#f = u^2, dg = e^(-u) du, df = 2 u du, g = -e^(-u): 2intu^2e^udu = -2 e^(-u) u^2 + 4 int ue^(-u) du#

For the integrand `ue^(-u)`, integrate by parts, `int f dg = f g - int g df`, where
#f = u, dg = e^(-u) du, df = du, g = -e^(-u): -2 e^(-u) u^2 + 4 int ue^(-u) du = -4 e^(-u) u - 2 e^(-u) u^2 + 4 int e^(-u) du#

The integral of `e^(-u)` is -`e^(-u):`

`-2 e^(-u) u^2 + 4 int ue^(-u) du= -2 e^(-u) u^2 - 4 e^(-u) u - 4 e^(-u) + c`

Substitute back for #u = sqrt(x): = -2 e^(-sqrt(x)) x - 4 e^(-sqrt(x)) sqrt(x) - 4 e^(-sqrt(x)) + c#

Which is equal to:

Answer: `= -2 e^(-sqrt(x)) (x + 2 sqrt(x) + 2) + c`

Answer 2

`intsqrtx(e^-sqrtx)dx=-2e^-sqrtx(x+2sqrtx+2)+C`

Explanation:

We'll start this integration by using substitution.
For `intsqrtx(e^-sqrtx)dx`, let `u=-sqrtx=-x^(1/2)`.
Therefore, `(du)/dx[-x^(1/2)]=-d/dx[x^(1/2)]=-(1/2x^(-1/2))=-1/(2sqrtx)`, so `du=-1/(2sqrtx)dx`.
To substitute we need something that's part of the integrand, so rearrange this to `-2sqrtxdu=dx`

Now our integral looks like `intsqrtx(e^u)(-2sqrtxdu)=int-2xe^udu`.
Remember that `u=-sqrtx`, therefore `u^2=x` so our integral is equivalent to `int-2u^2e^udu`, which we can solve.

First apply the constant rule, noting that `int-2u^2e^udu=-2color(green)(intu^2e^udu)`

Now we need to use integration by parts, which states that `intfdg=fg-intgdf`. Let `f=u^2` and `dg=e^udu`. `(df)/(du)=d/(du)[u^2]=2u`, so `df=2udu`. `intdg=inte^udu=e^u`, so `g=e^u`

`color(green)(intu^2e^udu)=u^2e^u-inte^u(2udu)=u^2e^u-2color(purple)(intue^udu)`

We need to use integration by parts again. This time, let `f=u` and `dg=e^udu`. `(df)/(du)=d/(du)[u]=1`, so `df=du`. `intdg=inte^udu=e^u`, so `g=e^u`

`color(purple)(intue^udu)=ue^u-inte^udu=ue^u-e^u`

`color(green)(intu^2e^udu)=u^2e^u-2color(purple)((ue^u-e^u))=u^2e^u-2ue^u+2e^u`

`-2(color(green)(u^2e^u-2ue^u+e^u))=-2u^2e^u+4ue^u-4e^u`

Finally, plug `-sqrtx` back in for `u` and simplify.

`-2xe^-sqrtx-4sqrtxe^-sqrtx-4e^-sqrtx=-2e^-sqrtx(x+2sqrtx+2)`

We've omitted the constant of integration thus far since all we were doing was integration by parts and substitution, so we'll put it in now.

`-2e^-sqrtx(x+2sqrtx+2)+C`