value of integral
∫[tex]xe^{x}[/tex]
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Solution:
Given Integral:
[tex]\displaystyle\tt\longrightarrow I=\int xe^x\:dx[/tex]
Now, we will use integral by parts to solve this problem.
Let us assume that:
[tex]\tt\longrightarrow u=x\:\:or\:\: du=dx[/tex]
[tex]\tt\longrightarrow dv=e^x\:dx\:\:or\:\: v=e^x[/tex]
So, using integral by parts, we get:
[tex]\displaystyle\tt\longrightarrow I=uv-\int v\:du[/tex]
[tex]\displaystyle\tt\longrightarrow I=xe^x-\int e^x\:dx[/tex]
[tex]\displaystyle\tt\longrightarrow I=xe^x-e^x+C[/tex]
[tex]\displaystyle\tt\longrightarrow I=e^x(x-1)+C[/tex]
Therefore:
[tex]\displaystyle\tt\longrightarrow \int xe^x\:dx=e^x(x-1)+C[/tex]
Which is our required answer.
Learn More:
[tex]\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}[/tex]
Consider the given Integral as
[tex]{:\implies \quad \displaystyle \bf I=\sf \int xe^{x}dx}[/tex]
So, here two functions are multiplied to form the so called Integrand we have, so we will use Integration by parts here, which is generally used to integrate the product of functions
So, Integrating by parts will yield
[tex]{:\implies \quad \displaystyle \bf I=\sf x\int e^{x}dx-\int \bigg(\dfrac{dx}{dx}\int e^{x}dx\bigg)dx}[/tex]
[tex]{:\implies \quad \displaystyle \bf I=\sf xe^{x}-\int e^{x}dx}[/tex]
[tex]{:\implies \quad \displaystyle \bf I=xe^{x}-e^{x}+C}[/tex]
Where, C is any Arbitrary Constant