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

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

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

area

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

\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{\partial}{\partial y} (2 y^{\frac{3}{2}})

\frac{y ^{^{'}} - e ^{- t} + 3}{y} = - 3, y (0) = 4

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (t)

\frac{d}{d x} (5 x cos (x))

x \to 0 + lim (3 - x)

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