- #1

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It looks obvious from the graph that the intersections are -1 and 1, but just to verify:

[tex]\frac{2}{{x^2 + 1}} = x^2 \,\, \Leftrightarrow \,\,x^2 \left( {x^2 + 1} \right) = 2\,\, \Leftrightarrow x^4 + x^2 = 2,\,\,x = - 1,\,\,1[/tex]

So my integration limits are -1 and 1

Therefore, my area should be

[tex]A = \int_{ - 1}^1 {\frac{2}{{x^2 + 1}}dx - \int_{ - 1}^1 {x^2 } } dx = 2\int_{ - 1}^1 {\frac{1}{{x^2 + 1}}dx - \int_{ - 1}^1 {x^2 } } dx[/tex]

Integrating this I get:

[tex]\left. {2\tan ^{ - 1} x - \frac{1}{3}x^3 } \right]_{ - 1}^1 = \left( {2\tan ^{ - 1} \left( 1 \right) - \frac{1}{3}1^3 } \right) - \left( {2\tan ^{ - 1} \left( { - 1} \right) - \frac{1}{3}\left( { - 1} \right)^3 } \right)[/tex]

Plugging this into my calculator gives me 1.90170379707065, but the back of the book says pi-2/3 which equals 2.475.

Just looking at the graph, it seems like the answer would be more than 2, as each of the 4 squares appear slightly more than half shaded.