What is $F (x) = \int e^{x - 2} - 3 x d x$ $F(x) = int e^(x-2) - 3x dx$ if $F (0) = 1$ $F(0) = 1$ ?

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What is $F (x) = \int e^{x - 2} - 3 x d x$ $F(x) = int e^(x-2) - 3x dx$ if $F (0) = 1$ $F(0) = 1$ ?

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Answer 1 · 2016-11-18T04:13:22

The function is $f (x) = e^{x - 2} - \frac{3}{2} {(x - 2)}^{2} - 6 (x - 2) - 5.135$ $f(x) = e^(x- 2) - 3/2(x- 2)^2 - 6(x- 2) - 5.135$ , approximately.

Explanation:

Let $u = x - 2$ $u = x-2$ , then $d u = d x$ $du = dx$ .

$\Rightarrow \int (e^{u} - 3 (u + 2)) d u$ $=>int(e^u - 3(u + 2))du$

$\Rightarrow \int (e^{u} - 3 u - 6) d u$ $=>int(e^u - 3u - 6)du$

$\Rightarrow e^{u} - \frac{3}{2} u^{2} - 6 u$ $=>e^u - 3/2u^2 - 6u$

We now substitute $u$ $u$ .

$\Rightarrow e^{x - 2} - \frac{3}{2} {(x - 2)}^{2} - 6 (x - 2) + C$ $=> e^(x - 2) - 3/2(x- 2)^2 - 6(x- 2) + C$

We know the input/output of the function so we can solve for $C$ $C$ .

When $x = 0$ $x = 0$ , $y = 1$ $y = 1$ .

$1 = e^{0 - 2} - \frac{3}{2} {(0 - 2)}^{2} - 6 (0 - 2) + C$ $1 = e^(0 - 2) - 3/2(0 - 2)^2 - 6(0 - 2) + C$

$1 = e^{- 2} - 6 + 12 + C$ $1 = e^(-2) - 6 + 12 + C$

$1 - \frac{1}{e^{2}} - 6 = C$ $1 - 1/e^2 - 6 = C$

$- 5 - \frac{1}{e^{2}} = C$ $-5 - 1/e^2 = C$

$C ≅ - 5.135$ $C~= -5.135$

Hopefully this helps!

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What is $F (x) = \int e^{x - 2} - 3 x d x$ $F(x) = int e^(x-2) - 3x dx$ if $F (0) = 1$ $F(0) = 1$ ?

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Explanation:

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