Solved: Inverse Functions: Triangle Area in First Quadrant

moela
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[SOLVED] Inverse Functions

What is the area of the largest triangle in the first quadrant with 2 sides on the axes and third side tangent to the curve y=e^-x
 
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just take any value of x and find the equation of the tangent line at that point, and see where it hits the x and y axes. then write a formula for that area, and maximize the formula by taking derivatives. hmmmmm... i think i got 2/e what did you get??
 
yea i got it
thanks

delete now please
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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