What is area f(x)=x^3-6x^2+8x,g(x)=x^2-4x ?

The answer to area f(x)=x^3-6x^2+8x,g(x)=x^2-4x is 71/6

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area f(x)=x^3-6x^2+8x,g(x)=x^2-4x

area

f (x) = x^{3} - 6 x^{2} + 8 x, g (x) = x^{2} - 4 x

\frac{71}{6}

Decimal

11.83333 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} x^{3} - 6 x^{2} + 8 x - (x^{2} - 4 x) d x

Solve

\int_{0}^{4} x^{3} - 6 x^{2} + 8 x - (x^{2} - 4 x) d x : \frac{71}{6}

The area is:

= \frac{71}{6}

\frac{1}{( x - 7 )}

y^{^{'}} - y = e^{t} cos (t), y (0) = 1

\frac{d}{d x} (\frac{5}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}})

\frac{d}{d x} (sin (2 x) - 2 sin (x))

\int cos (x) cot (x) d x

What is area f(x)=x^3-6x^2+8x,g(x)=x^2-4x ?
The answer to area f(x)=x^3-6x^2+8x,g(x)=x^2-4x is 71/6