What is area f(x)= 1/(4(1+x^2)),g(x)= 1/8 x^2 ?

The answer to area f(x)= 1/(4(1+x^2)),g(x)= 1/8 x^2 is pi/8-1/12

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area f(x)= 1/(4(1+x^2)),g(x)= 1/8 x^2

area

f (x) = \frac{1}{4 ( 1 + x ^{2} )}, g (x) = \frac{1}{8} x^{2}

\frac{π}{8} - \frac{1}{12}

Decimal Notation

0.30936 \dots

Solution steps

Apply the area formula:

\int_{- 1}^{1} \frac{1}{4 ( 1 + x ^{2} )} - \frac{1}{8} x^{2} d x

Solve

\int_{- 1}^{1} \frac{1}{4 ( 1 + x ^{2} )} - \frac{1}{8} x^{2} d x : \frac{π}{8} - \frac{1}{12}

The area is:

= \frac{π}{8} - \frac{1}{12}

\frac{d}{d x} (\frac{2 y}{x ^{2} + y ^{2}})

x \to 45 lim (\frac{x ^{2}}{9} - \frac{5}{x})

\frac{d}{d x} (sin^{5} (2 x))

\frac{\partial}{\partial x} (x e^{x^{2}})

x y^{^{'}} - x^{2} y = - x^{2}

What is area f(x)= 1/(4(1+x^2)),g(x)= 1/8 x^2 ?
The answer to area f(x)= 1/(4(1+x^2)),g(x)= 1/8 x^2 is pi/8-1/12