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How do you integrate #int x(5^(-x^2))dx#?

Answer 1

Samantha Adkison

The answer is #=-(1/2)5^(-x^2)/ln5+C#

We use the substitution

#u=-x^2#

#du=-2xdx#

#xdx=(-du)/2#

Therefore,

#intx(5^(-x^2))dx=-1/2int 5^(u)du#

Let, #y=5^(u)#

Then taking the logarithm

#lny=u ln5#

#y=e^(u ln5)#

#int 5^(u)du=inte^(u ln5)du=e^(u ln5)/ln5=y/ln5=5^(u)/ln5#

Therefore,

#intx(5^(-x^2))dx=-(1/2)5^(-x^2)/ln5+C#

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Answer 2

Christopher Allers

To integrate (\int x \cdot 5^{-x^2} , dx), we can use the technique of substitution. Let (u = -x^2), then (du = -2x , dx). Solving for (dx), we get (dx = -\frac{du}{2x}). Substituting these into the integral, we have (\int x \cdot 5^{-x^2} , dx = -\frac{1}{2} \int e^u , du). Now, integrate (-\frac{1}{2} \int e^u , du) with respect to (u). After integrating, substitute back (u = -x^2) to obtain the final answer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.