# AREA under the curve

• Nov 27th 2010, 06:16 AM
pankaj
AREA under the curve
1) For $\displaystyle x\in[-\sqrt{2},1),$the equation $\displaystyle (y-1)x^2-3(y+1)x+2(y-1)=0$ defines a monotonic function $\displaystyle y=f(x).$Find area bounded by $\displaystyle y=f^{-1}(x),x=-17+12\sqrt{2},x=1$ and the x-axis.

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

3) If $\displaystyle f(x)=\sin x,\forall x\in[0,\frac{\pi}{2}],$$\displaystyle 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 $\displaystyle y=f(x)$ and the x-axis.
• Nov 27th 2010, 10:39 PM
CaptainBlack
Quote:

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

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

CB