Respuesta :
Answer:
[tex]-42.79587[/tex].
Step-by-step explanation:
We have been given the indefinite integral [tex]\int _1^2\:\left(1-x^2\right)e^{2x}\:dx[/tex]. We are asked to find the integral using integration by parts.
We will use Integration by parts formula to solve our given problem.
[tex]\int\ vdv=uv-\int\ vdu[/tex]
Let [tex]u=1-x^2[/tex] and [tex]v'=e^{2x}[/tex].
Now, we need to find du and v using these values as shown below:
[tex]\frac{du}{dx}=\frac{d}{dx}(1-x^2)[/tex]
[tex]\frac{du}{dx}=-2x[/tex]
[tex]du=-2xdx[/tex]
[tex]v'=e^{2x}[/tex]
[tex]v=\frac{1}{2}e^{2x}[/tex]
Upon substituting these values in integration by parts formula, we will get:
[tex]\int _1^2\:\left(1-x^2\right)e^{2x}\:dx=(1-x^2)(\frac{e^{2x}}{2})-\int _1^2(\frac{e^{2x}}{2})(-2x)dx[/tex]
[tex]\int _1^2\:\left(1-x^2\right)e^{2x}\:dx=(1-x^2)(\frac{e^{2x}}{2})-\int _1^2 -xe^{2x}dx[/tex]
[tex]\int _1^2\:\left(1-x^2\right)e^{2x}\:dx=(1-x^2)(\frac{e^{2x}}{2})+\frac{1}{4}(e^{2x}(2x)-e^{2x})[/tex]
Let us compute the boundaries.
[tex](1-2^2)(\frac{e^{2*2}}{2})+\frac{1}{4}(e^{2*2}(2*2)-e^{2*2})=-\frac{3e^4}{4}=-40.94861[/tex]
[tex](1-1^2)(\frac{e^{2*1}}{2})+\frac{1}{4}(e^{2*1}(2*1)-e^{2*1})=\frac{e^2}{4}=1.84726[/tex]
[tex]-40.94861-1.84726=-42.79587[/tex]
Therefore, the value pf the given definite integral would be [tex]-42.79587[/tex].
