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

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

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Answer 1 · 2017-07-10T03:10:42

Zero.

Explanation:

First integrate:

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

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

So, for $f (1) = 0$ $f(1)=0$ , we have:

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

Then you can solve for C.

$C = - \frac{e^{2}}{2} + e - \frac{1}{3}$ $C=-e^2/2+e-1/3$

So,

$f (x) = \int e^{2 x} - e^{x} + x^{2} d x = \frac{1}{2} e^{2} - e + \frac{1}{3} + (- \frac{e^{2}}{2} + e - \frac{1}{3})$ $f(x)=inte^(2x)-e^x+x^2dx=1/2e^2-e+1/3+(-e^2/2+e-1/3)$

$= 0$ $=0$

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

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What is f(x) = int e^(2x)-e^x+x^2 dxf(x)=∫e2x−ex+x2dxf(x) = int e^(2x)-e^x+x^2 dx if f(1) = 0 f(1)=0f(1) = 0 ?

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