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

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T08:46:29

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

$\int 3 x^{2} + e^{2 - 2 x} d x = 3 \int x^{2} d x + e^{2} \int e^{- 2 x} d x$ $int 3x^2+e^(2-2x) dx = 3intx^2 dx+ e^2 int e^(-2x) dx$

$= 3 \cdot \frac{x^{3}}{3} + e^{2} \int e^{- 2 x} d x$ $= 3*x^3/3 + e^2 int e^(-2x) dx$

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

$:. F(x) = x^3 - e^2/2 * e^(-2x) +C$

Since $F(0) =1$

$0^3 - e^2/2 * e^(-2*0) +C =1$

$-e^2/2 *1 +C =1$

$C= 1+e^2/2$

$F(x) = x^3 - e^2/2 * e^(-2x) +1+e^2/2$

$= x^3 +e^2/2 (1-e^(-2x)) +1$

What is F(x) = int 3x^2+e^(2-2x) dxF(x)=∫3x2+e2−2xdxF(x) = int 3x^2+e^(2-2x) dx if F(0) = 1 F(0)=1F(0) = 1 ?