What is area f(x)=x^2-4x,g(x)=0 ?

The answer to area f(x)=x^2-4x,g(x)=0 is 32/3

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area f(x)=x^2-4x,g(x)=0

area

f (x) = x^{2} - 4 x, g (x) = 0

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

Apply the area formula:

\int_{0}^{4} x^{2} - 4 x d x

Solve

\int_{0}^{4} x^{2} - 4 x d x : \frac{32}{3}

The area is:

= \frac{32}{3}

\frac{d}{d x} ((5 x^{7} - 7 x^{5} + 12) (3 x^{6} - 2 x))

x \to 1 lim (x + 3)

(x^{3} + 2)^{5}

\frac{\partial}{\partial x} (arctan (x))

\int x^{3} I n x d x