What is area y= 4/(1+x^2),y=1 ?

The answer to area y= 4/(1+x^2),y=1 is (8pi)/3-2sqrt(3)

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area y= 4/(1+x^2),y=1

area

y = \frac{4}{1 + x ^{2}}, y = 1

\frac{8 π}{3} - 23

Decimal Notation

4.91347 \dots

Solution steps

Apply the area formula:

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

Solve

\int_{- 3}^{3} \frac{4}{1 + x ^{2}} - 1 d x : \frac{8 π}{3} - 23

The area is:

= \frac{8 π}{3} - 23

\frac{d}{d t} (t^{2} - 2 t)

x^{^{'}} = 0.1078 \cdot (1 - \frac{x}{105900}) x - 0.1 \cdot x

(2.6) (5.15)

x \to 1 lim (2 x^{3} - 4 x^{2} + x + 9)

5 x^{2} + x - 5, at x = 8

What is area y= 4/(1+x^2),y=1 ?
The answer to area y= 4/(1+x^2),y=1 is (8pi)/3-2sqrt(3)