Region R is enclosed by the graphs of y=tanx^2, y=(1/2)secx^2. Find R. Take it from [0,1] For the integral I got (1/2)tanx-tanx-x. Taking the integral from [0,1] I got about .22129. Is that right?
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Originally Posted by ment2byours Region R is enclosed by the graphs of y=tanx^2, y=(1/2)secx^2. Find R. Take it from [0,1] For the integral I got (1/2)tanx-tanx-x. Taking the integral from [0,1] I got about .22129. Is that right? you are aware that the curves intersect at $\displaystyle \frac{\pi}{4}$ , right? $\displaystyle A = \int_0^{\frac{\pi}{4}} \frac{\sec^2{x}}{2} - \tan^2{x} \, dx + \int_{\frac{\pi}{4}}^1 \tan^2{x} - \frac{\sec^2{x}}{2} \, dx$ $\displaystyle A = .3495$
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