Aspire Faculty ID #16535 · Topic: NIMCET 2008 · Just now
NIMCET 2008

The area of the plane bounded by $y=\sqrt{x},\ x\in[0,1]$, $y=x^2,\ x\in[1,2]$, $y=-x^2+2x+4,\ x\in[0,2]$

Solution

Area = $\displaystyle \int_0^1[( -x^2+2x+4)-\sqrt{x}]dx + \int_1^2[( -x^2+2x+4)-x^2]dx$ Evaluating gives $\boxed{\frac{19}{3}}$

Previous 10 Questions — NIMCET 2008

Nearest first

Next 10 Questions — NIMCET 2008

Ascending by ID
Ask Your Question or Put Your Review.

loading...