(d) $Required\ area = \int_{0}^{2}{y\ \ dx}$ [image:8_image46.jpeg] $= \int_{0}^{2}{\frac{x^{2}}{2}dx = \left\lbrack \frac{x^{3}}{6} \right\rbrack_{0}^{2} = \frac{4}{3}\ sq\ units}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.