What is area y=-3x^3+x^2+10x,x>= 0 ?

The answer to area y=-3x^3+x^2+10x,x>= 0 is diverges

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area y=-3x^3+x^2+10x,x>= 0

area

y = - 3 x^{3} + x^{2} + 10 x, x \geq 0

diverges

Solution steps

Apply the area formula:

\int_{0}^{\infty} - 3 x^{3} + x^{2} + 10 x d x

Solve

\int_{0}^{\infty} - 3 x^{3} + x^{2} + 10 x d x :

diverges

The area is:

= diverges

y^{^{''}} - 2 y^{^{'}} - 3 y = x e^{x}

e^{\frac{4}{5} x}

\frac{d}{d x} (x + 1 - 2)

\frac{d}{d x} (x^{2} - 32 x)

\frac{d}{d x} (\frac{e ^{4 x^{2 + 1}}}{8})

What is area y=-3x^3+x^2+10x,x>= 0 ?
The answer to area y=-3x^3+x^2+10x,x>= 0 is diverges