What is area f(x)=x^2-4,g(x)=0,x=-4,x=3 ?

The answer to area f(x)=x^2-4,g(x)=0,x=-4,x=3 is 71/3

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area f(x)=x^2-4,g(x)=0,x=-4,x=3

area

f (x) = x^{2} - 4, g (x) = 0, x = - 4, x = 3

\frac{71}{3}

Decimal Notation

23.66666 \dots

Solution steps

Apply the area formula:

\int_{- 4}^{3} x^{2} - 4 d x

Solve

\int_{- 4}^{3} x^{2} - 4 d x : \frac{71}{3}

The area is:

= \frac{71}{3}

\frac{\partial}{\partial x} (2 x - cos (y))

\frac{d}{d x} (\frac{sin ( x )}{1 - sin ( x )})

\frac{\partial}{\partial y} (25 - x^{2} - 5 y)

\int 6 x 1 + x^{2} d x

y^{^{''}} + 20 y^{^{'}} + 500 y = 0

What is area f(x)=x^2-4,g(x)=0,x=-4,x=3 ?
The answer to area f(x)=x^2-4,g(x)=0,x=-4,x=3 is 71/3