What is #f(x) = int xe^(2x) + x^3 dx# if #f(-3 ) = 1 #?

Question

1523 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T19:50:23

# f(x)=1/4{(2x-1)e^(2x)+x^4+7(e^-6-11)}#.

#f(x)=int(xe^(2x)+x^3)dx#,

#=intxe^(2x)dx+intx^3dx#,

#=intxe^(2x)dx+x^4/4#,

#=[x*inte^(2x)dx-int(d/dx(x)inte^(2x)dx)dx]+x^4/4...[because, IBP]#,

#=x*e^(2x)/2-int(1*e^(2x)/2)dx+x^4/4#,

# rArr f(x)=(xe^(2x))/2-1/2*e^(2x)/2+x^4/4+C............(ast)#.

To determine #C#, we use the given condition : #f(-3)=1#.

#f(-3)=1, &, (ast) rArr 1=(-3*e^-6)/2-e^-6/4+81/4+C#.

#:. C=7/4(e^-6-11)#.

# rArr f(x)=1/4{(2x-1)e^(2x)+x^4+7(e^-6-11)}#.

Enjoy Maths.!