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Question: What is the integration of the given equation \(\left( {{x^2}} \right)\,\left( {{e^x}} \right)\,dx\,...

What is the integration of the given equation (x2) (ex) dx  ?\left( {{x^2}} \right)\,\left( {{e^x}} \right)\,dx\,\,?

Explanation

Solution

For solving this equation, first of all we should be aware of Integration, its uses and applications. Integration is just the calculation of an integral. We can also define the integration as the technique for finding a function g(x)g\left( x \right) which is the derivative of the function d f(x)d\,f\left( x \right).
Types of Integration:-
There are two types of Integration, which includes Definite Integration and Indefinite Integration.
1. Definite Integral - It generally represents a number with upper and lower limits as constant.
2. Indefinite Integral - It generally represents a number with upper and lower limits as variable.

Complete step by step answer:
From the question, now we will integrate the value (x2) (ex) dx  \left( {{x^2}} \right)\,\left( {{e^x}} \right)\,dx\,\,
Let u=x2  and   v=ex    then,    du=2xdx       and       dv=exdxu = {x^2}\,\,and\,\,\,v = {e^x}\,\,\,\,then,\,\,\,\,du = 2xdx\,\,\,\,\,\,\,and\,\,\,\,\,\,\,dv = {e^x}dx

Now by using integration by parts:-
⇒∫u(x)v1 (x)dx=   u(x)v(x)  −  ∫v(x)u1(x)dx\Rightarrow \int {u\left( x \right)} {v^1}\,\left( x \right)dx = \,\,\,u\left( x \right)v\left( x \right)\,\, - \,\,\int {v\left( x \right){u^1}\left( x \right)dx}
Hence, ∫x2exdx=  x2ex−  ∫ex ×  2xdx\int {{x^2}{e^x}dx = \,\,{x^2}{e^x} - } \,\,\int {{e^x}\, \times \,\,2xdx}
⇒x2ex− 2∫xexdx +  c   ............−(i)\Rightarrow {x^2}{e^x} - \,2\int {x{e^x}dx\, + \,\,c} \,\,\,............ - \left( i \right)

Now by putting u=x      then     du=dxu = x\,\,\,\,\,\,then\,\,\,\,\,du = dx
And ∫xexdx= xex−  ∫ex × 1  ×  dx\int {x{e^x}dx = \,x{e^x} - \,\,\int {{e^x}\, \times \,1\,\, \times \,\,dx} } or ∫xexdx= xex−  ∫exdx=  xe−ex..........−(ii)\int{x{e^x}dx = \,x{e^x} - \,\,\int {{e^x}dx = \,\,{x^e} - {e^x}} } .......... - \left( {ii} \right)
Hence, substituting the value of equation (ii) in (i) we get:-
⇒∫x2exdx=x2ex−2(xex−ex)+c\Rightarrow \int {{x^2}{e^x}dx} = {x^2}{e^x} - 2\left( {x{e^x} - {e^x}} \right) + c
⇒ex(x2−2x+2)  +  c\Rightarrow {e^x}\left( {{x^2} - 2x + 2} \right)\,\, + \,\,c
Hence, the required value of the equation (x2) (ex) dx  \left( {{x^2}} \right)\,\left( {{e^x}} \right)\,dx\,\, is ex(x2−2x+2)  +  c{e^x}\left( {{x^2} - 2x + 2} \right)\,\, + \,\,c

Note: The principles of integration were given by Isaac Newton and Gottfried Wilhelm Leibniz in the late seventh century. Gottfried Wilhelm Leibniz published his work on calculus before Isaac Newton. Integration is also known as anti-differentiation.