AREA under the curve

**pankaj** · November 27th 2010, 06:16 AM

1) For $x\in[-\sqrt{2},1),$ the equation $(y-1)x^2-3(y+1)x+2(y-1)=0$ defines a monotonic function $y=f(x).$ Find area bounded by $y=f^{-1}(x),x=-17+12\sqrt{2},x=1$ and the x-axis.

2) Let $f(x)$ be continuous and bijective such that $f(0)=0$ .If $\forall t\in R$ ,area bounded by $y=f(x),x=a-t,x=a$ and x-axis is equal to area bounded by $y=f(x),x=a+t,x=a$ and x-axis,then prove that
$\int_{-\lambda}^{\lambda} f^{-1}(x)=2a\lambda$

3) If $f(x)=\sin x,\forall x\in[0,\frac{\pi}{2}],$ $f(x)+f(\pi-x)=2 \forall x\in\big(\frac{\pi}{2},\pi], f(x)=f(2\pi-x),\forall x\in\big(\frac{3\pi}{2},2\pi\big]$ ,then find the area enclosed by $y=f(x)$ and the x-axis.

**CaptainBlack** · November 27th 2010, 10:39 PM

Originally Posted by pankaj

1) For $x\in[-\sqrt{2},1),$ the equation $(y-1)x^2-3(y+1)x+2(y-1)=0$ defines a monotonic function $y=f(x).$ Find area bounded by $y=f^{-1}(x),x=-17+12\sqrt{2},x=1$ and the x-axis.

Rewrite the equation in the form $y=f(x)$ , sketch the curve and on that sketch shade the area definded. This will allow you to express the area as an integral of $f(x)$ .

CB

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