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

Question

1221 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-12T17:03:18

#f(x)=3/4x^4+(x-1)e^(x-2)-9-e.#

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

#=int3x^3dx+intxe^(x-2)dx#

#=3(x^(3+1)/(3+1))+I, where, I=intxe^(x-2)dx,#

# :. f(x)=3/4x^4+I,................(ast).#

To evaluate #I,# we use the following Method of Integration

by Parts (IBP).

# (IBP) : intuvdx=uintvdx-int[(du)/dx*intvdx]dx.#

We take, #u=x rArr (du)/dx=1, &, v=e^(x-2) rArr intvdx=e^(x-2).#

# :. I=xe^(x-2)-int{1*e^(x-2)}dx,#

#:. I=xe^(x-2)-e^(x-2)=(x-1)e^(x-2).............(star).#

#(ast) and (star) rArr f(x)=3/4x^4+(x-1)e^(x-2)+C.#

To determine #C," we use the given cond. : "f(2)=3.#

# rArr 3/4(2)^4+(2-1)e^(3-2)+C=3 rArr C=-9-e#

Therefore, #f(x)=3/4x^4+(x-1)e^(x-2)-9-e.#

Enjoy Maths.!