What is $f (x) = \int e^{2 x} - e^{x} + x d x$ $f(x) = int e^(2x)-e^x+x dx$ if $f (4) = 2$ $f(4 ) = 2$ ?

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What is $f (x) = \int e^{2 x} - e^{x} + x d x$ $f(x) = int e^(2x)-e^x+x dx$ if $f (4) = 2$ $f(4 ) = 2$ ?

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Answer 1 · 2016-07-04T17:28:40

$f (x) = \frac{e^{2 x}}{2} - e^{x} + \frac{x^{2}}{2} + e^{4} - \frac{e^{8}}{2} - 6$ $f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6$

Explanation:

Let $I = \int (e^{2 x} - e^{x} + x) d x = \frac{e^{2 x}}{2} - e^{x} + \frac{x^{2}}{2} + C$ $I=int(e^(2x)-e^x+x)dx=e^(2x)/2-e^x+x^2/2+C$ , $C$ $C$ is a constant of integration.

To determine $C$ $C$ , we are given the cond. that $f (4) = 2$ $f(4)=2$ .

Now, $f (x) = I = \frac{e^{2 x}}{2} - e^{x} + \frac{x^{2}}{2} + C .$ $f(x)=I=e^(2x)/2-e^x+x^2/2+C.$ Hence,
$f (4) = 2 \Rightarrow \frac{e^{8}}{2} - e^{4} + 8 + C = 2 \Rightarrow C = e^{4} - \frac{e^{8}}{2} - 6 .$ $f(4)=2 rArr e^8/2-e^4+8+C=2 rArr C=e^4-e^8/2-6.$

Sub.ing this value of $C$ $C$ in $I$ $I$ , we get,

$f (x) = \frac{e^{2 x}}{2} - e^{x} + \frac{x^{2}}{2} + e^{4} - \frac{e^{8}}{2} - 6$ $f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6$

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What is $f (x) = \int e^{2 x} - e^{x} + x d x$ $f(x) = int e^(2x)-e^x+x dx$ if $f (4) = 2$ $f(4 ) = 2$ ?

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Explanation:

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What is $f (x) = \int e^{2 x} - e^{x} + x d x$ $f(x) = int e^(2x)-e^x+x dx$ if $f (4) = 2$ $f(4 ) = 2$ ?