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

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

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

area

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

f (x) = \frac{x ^{2} - 1}{2 x - 3}

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0

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

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

x \to 1 lim (\frac{ln ( x )}{x})

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