What is area y=8x^2-2x,y=x^3-8x+3 ?

The answer to area y=8x^2-2x,y=x^3-8x+3 is 232sqrt(69)+(-3291sqrt(69)-6217)/(24)

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area y=8x^2-2x,y=x^3-8x+3

area

y = 8 x^{2} - 2 x, y = x^{3} - 8 x + 3

23269 + \frac{- 3291 69 - 6217}{24}

Decimal

529.04927 \dots

Solution steps

Apply the area formula:

\int_{- 1}^{\frac{9 + 69}{2}} 8 x^{2} - 2 x - (x^{3} - 8 x + 3) d x

Solve

\int_{- 1}^{\frac{9 + 69}{2}} 8 x^{2} - 2 x - (x^{3} - 8 x + 3) d x : 23269 + \frac{- 3291 69 - 6217}{24}

The area is:

= 23269 + \frac{- 3291 69 - 6217}{24}

\frac{d y}{d x} = \frac{y ^{2} + x y + x ^{2}}{x ^{2}}

\int sin^{2} (y) cos (y) d y

\frac{1}{( s - 1 ) ( s + 4 )}

\int (4 x^{3} - 2 x) (x^{4} - x^{2} - 5)^{3} d x

x \to + 0 + lim (arctan (\frac{2}{x}))

What is area y=8x^2-2x,y=x^3-8x+3 ?
The answer to area y=8x^2-2x,y=x^3-8x+3 is 232sqrt(69)+(-3291sqrt(69)-6217)/(24)