What is $f (x) = \int 3 x^{3} + x e^{x - 2} d x$ $f(x) = int 3x^3+xe^(x-2) dx$ if $f (2) = 3$ $f(2 ) = 3$ ?

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Answer 1 · 2017-06-12T17:03:18

$f (x) = \frac{3}{4} x^{4} + (x - 1) e^{x - 2} - 9 - e .$ $f(x)=3/4x^4+(x-1)e^(x-2)-9-e.$

$f (x) = \int (3 x^{3} + x e^{x - 2}) d x .$ $f(x)=int(3x^3+xe^(x-2))dx.$

$= \int 3 x^{3} d x + \int x e^{x - 2} d x$ $=int3x^3dx+intxe^(x-2)dx$

$= 3 (\frac{x^{3 + 1}}{3 + 1}) + I, w h e r e, I = \int x e^{x - 2} d x,$ $=3(x^(3+1)/(3+1))+I, where, I=intxe^(x-2)dx,$

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

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.!

What is f(x) = int 3x^3+xe^(x-2) dxf(x)=∫3x3+xex−2dxf(x) = int 3x^3+xe^(x-2) dx if f(2 ) = 3 f(2)=3f(2 ) = 3 ?