What is area f(x)=x^2+4,g(x)=-x^2-1,[-2,3] ?

The answer to area f(x)=x^2+4,g(x)=-x^2-1,[-2,3] is 145/3

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area f(x)=x^2+4,g(x)=-x^2-1,[-2,3]

area

f (x) = x^{2} + 4, g (x) = - x^{2} - 1, [- 2, 3]

\frac{145}{3}

Decimal Notation

48.33333 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 2}^{3} x^{2} + 4 - (- x^{2} - 1) d x : \frac{145}{3}

The area is:

= \frac{145}{3}

x \to 2 lim (\frac{x - 3}{x + 5})

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 2, y (0) = 1

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

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

n = 0 \sum \infty 9 (\frac{x - 3}{4})^{n}

What is area f(x)=x^2+4,g(x)=-x^2-1,[-2,3] ?
The answer to area f(x)=x^2+4,g(x)=-x^2-1,[-2,3] is 145/3