What is area y= 9/(x^2),y=10-x^2,y=12 ?

The answer to area y= 9/(x^2),y=10-x^2,y=12 is (8(9sqrt(3)-10))/3

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area y= 9/(x^2),y=10-x^2,y=12

area

y = \frac{9}{x ^{2}}, y = 10 - x^{2}, y = 12

\frac{8 ( 9 3 - 10 )}{3}

Decimal Notation

14.90255 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{9}{x ^{2}} = 10 - x^{2} : x = 1, x = - 1, x = 3, x = - 3

\frac{9}{x ^{2}} = 12 : x = \frac{3}{2}, x = - \frac{3}{2}

10 - x^{2} = 12 :

No Solution for

x \in R

= \int_{- 3}^{- 1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- 1}^{- \frac{3}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- \frac{3}{2}}^{\frac{3}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{3}{2}}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{3} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 3}^{- 1} \frac{9}{x ^{2}} - (10 - x^{2}) d x + \int_{- 1}^{- \frac{3}{2}} \frac{9}{x ^{2}} - (10 - x^{2}) d x + \int_{- \frac{3}{2}}^{\frac{3}{2}} 10 - x^{2} - 12 d x + \int_{\frac{3}{2}}^{1} \frac{9}{x ^{2}} - (10 - x^{2}) d x + \int_{1}^{3} \frac{9}{x ^{2}} - (10 - x^{2}) d x

\int_{- 3}^{- 1} \frac{9}{x ^{2}} - (10 - x^{2}) d x = \frac{16}{3}

\int_{- 1}^{- \frac{3}{2}} \frac{9}{x ^{2}} - (10 - x^{2}) d x = \frac{261 3 - 448}{24}

\int_{- \frac{3}{2}}^{\frac{3}{2}} 10 - x^{2} - 12 d x = \frac{3}{4} + 23

\int_{\frac{3}{2}}^{1} \frac{9}{x ^{2}} - (10 - x^{2}) d x = \frac{261 3 - 448}{24}

\int_{1}^{3} \frac{9}{x ^{2}} - (10 - x^{2}) d x = \frac{16}{3}

= \frac{16}{3} + \frac{261 3 - 448}{24} + \frac{3}{4} + 23 + \frac{261 3 - 448}{24} + \frac{16}{3}

Simplify

= \frac{8 ( 9 3 - 10 )}{3}

\frac{d y}{d x} - \frac{y}{2 x} = 3 x

x (x^{2} - 11 x + 35), x^{2}, 0, 7

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

\int_{x}^{0} sin (t^{2}) d t

What is area y= 9/(x^2),y=10-x^2,y=12 ?
The answer to area y= 9/(x^2),y=10-x^2,y=12 is (8(9sqrt(3)-10))/3