What is $f (x) = \int x e^{x} - 5 x d x$ $f(x) = int xe^x-5x dx$ if $f (- 1) = 5$ $f(-1) = 5$ ?

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Answer 1 · 2016-06-04T02:21:25

$f (x) = x e^{x} - e^{x} - \frac{5}{2} x^{2} + 2 e^{- 1} + \frac{15}{2}$ $f(x)=xe^x-e^x-5/2x^2+2e^(-1)+15/2$

The given

$\int (x e^{x} - 5 x) d x$ $int (xe^x-5x) dx$ and $f (- 1) = 5$ $f(-1)=5$

Integration by parts for the $\int (x e^{x}) d x$ $int (xe^x) dx$

Let $u = x$ $u=x$ and $d v = e^{x} \cdot d x$ $dv=e^x*dx$ and $v = e^{x}$ $v=e^x$ and $d u = d x$ $du=dx$

Using the integration by parts formula

$\int u \cdot d v = u v - \int v \cdot d u$ $int u*dv=uv-int v*du$

$\int x \cdot e^{x} \cdot d x = x e^{x} - \int e^{x} \cdot d x$ $int x*e^x*dx=xe^x-int e^x*dx$

$\int x \cdot e^{x} \cdot d x = x e^{x} - e^{x}$ $int x*e^x*dx=xe^x- e^x$

Therefore

$f (x) = \int (x e^{x} - 5 x) d x = x e^{x} - e^{x} - \frac{5}{2} x^{2} + C$ $f(x)=int (xe^x-5x) dx=xe^x- e^x-5/2x^2+C$

$f (x) = x e^{x} - e^{x} - \frac{5}{2} x^{2} + C$ $f(x)=xe^x- e^x-5/2x^2+C$

At $f (- 1) = 5$ $f(-1)=5$

$f (- 1) = 5 = - 1 \cdot e^{- 1} - e^{- 1} - \frac{5}{2} {(- 1)}^{2} + C$ $f(-1)=5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C$

$5 = - 1 \cdot e^{- 1} - e^{- 1} - \frac{5}{2} {(- 1)}^{2} + C$ $5=-1*e^(-1)- e^(-1)-5/2(-1)^2+C$

$5 = - 2 \cdot e^{- 1} - \frac{5}{2} + C$ $5=-2*e^(-1)-5/2+C$

$C = \frac{15}{2} + 2 e^{- 1}$ $C=15/2+2e^(-1)$

finally

$f (x) = x e^{x} - e^{x} - \frac{5}{2} x^{2} + \frac{15}{2} + 2 e^{- 1}$ $f(x)=xe^x- e^x-5/2x^2+15/2+2e^(-1)$

God bless....I hope the explanation is useful.

What is f(x) = int xe^x-5x dxf(x)=∫xex−5xdxf(x) = int xe^x-5x dx if f(-1) = 5 f(−1)=5f(-1) = 5 ?