What is #f(x) = int xe^(x^2-1)+2x dx# if #f(0) = -4 #?

Question

What is #f(x) = int xe^(x^2-1)+2x    dx# if #f(0) = -4  #?

Luke Alvarez · Accepted Answer

#f(x)=1/2e^(x^2-1)+x^2-4-1/(2e)#

Split up the integral:

#f(x)=intxe^(x^2-1)dx+int2xdx#

For the first integral, use substitution: let #u=x^2-1#, which implies that #du=2xdx#.

Multiply the integrand by #2# and the exterior of the integral by #1/2#.

#f(x)=1/2int2xe^(x^2-1)dx+int2xdx#

Substituting in #u=x^2-1# and #du=2xdx#:

#f(x)=1/2inte^udu+2intxdx#

Note that #inte^udu=e^u+C#. We will refrain from adding the constant of integration until we evaluate the second integral.

#f(x)=1/2e^u+2intxdx#

Since #u=x^2-1#:

#f(x)=1/2e^(x^2-1)+2intxdx#

In order to integrate #2intxdx#, use the rule #intx^ndx=x^(n+1)/(n+1)+C#, where #n!=-1#.

Applying this rule:

#f(x)=1/2e^(x^2-1)+2((x^(1+1))/(1+1))+C#

#f(x)=1/2e^(x^2-1)+x^2+C#

Now, we can determine the value of #C# using the initial condition #f(0)=-4#.

#-4=1/2e^(0^2-1)+0^2+C#

#-4=1/2e^-1+C#

#-4-1/(2e)=C#

Hence:

#f(x)=1/2e^(x^2-1)+x^2-4-1/(2e)#

Elijah Abram · Answer

undefined To find $ f(x) $, integrate the given function with respect to $ x $, then use the provided initial condition $ f(0) = -4 $ to find the constant of integration.

$ f(x) = \int (xe^{x^2 - 1} + 2x) \, dx $

Integrating $ xe^{x^2 - 1} $ with respect to $ x $ gives:

$ \int xe^{x^2 - 1} \, dx = \frac{1}{2} e^{x^2 - 1} + C $

And integrating $ 2x $ with respect to $ x $ gives:

$ \int 2x \, dx = x^2 + C_2 $

Now, $ f(x) $ becomes:

$ f(x) = \frac{1}{2} e^{x^2 - 1} + x^2 + C $

Given $ f(0) = -4 $, we can substitute $ x = 0 $ into the equation:

$ -4 = \frac{1}{2} e^{0^2 - 1} + 0^2 + C $

$ -4 = \frac{1}{2} e^{-1} + C $

$ -4 + 2e^{-1} = C $

$ C = 2e^{-1} - 4 $

Therefore, the function $ f(x) $ is:

$ f(x) = \frac{1}{2} e^{x^2 - 1} + x^2 + 2e^{-1} - 4 $