What is $f (x) = \int 5 e^{2 x} - 2 e^{x} - x d x$ $f(x) = int 5e^(2x)-2e^x-x dx$ if $f (4) = 1$ $f(4 ) = 1$ ?

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Answer 1 · 2016-11-14T12:46:59

$f (x) = \frac{5}{2} e^{2 x} - 2 e^{x} - \frac{x^{2}}{2} + 9 - \frac{5}{2} e^{8} + 2 e^{4}$ $f(x) = 5/2e^(2x) - 2e^x - x^2/2 + 9 - 5/2e^8 + 2e^4$

We have $f (x) = \int 5 e^{2 x} - 2 e^{x} - x d x$ $f(x)=int5e^(2x)-2e^x-xdx$

Integrating gives us:

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

We know $f (4) = 1$ $f(4)=1$ , so

$\frac{5}{2} e^{8} - 2 e^{4} - \frac{16}{2} + C = 1$ $5/2e^8 - 2e^4 - 16/2 + C = 1$
$:. C = 9 - 5/2e^8 + 2e^4$

Hence,

$f(x) = 5/2e^(2x) - 2e^x - x^2/2 + 9 - 5/2e^8 + 2e^4$

What is f(x) = int 5e^(2x)-2e^x-x dxf(x)=∫5e2x−2ex−xdxf(x) = int 5e^(2x)-2e^x-x dx if f(4 ) = 1 f(4)=1f(4 ) = 1 ?