Question

Find the area of the region in the first quadrant bounded by the parabola y = 4x² and the lines x = 0, y = 1 and y = 4.

Answer 1

\[y = 4 x^2\text{ represents a parabola , openeing upwards, symmetrical about + ve } y - \text{ axis and having vertex at O}(0, 0)\]
\[y = 1\text{ is a line parallel to }x - \text{ axis , cutting parabola at }\left( - \frac{1}{2}, 1 \right) and \left( \frac{1}{2}, 1 \right)\]
\[y = 4\text{ is a line parallel to } x \text{ axis , cutting parabola at }\left( - 1, 1 \right)\text{ and }\left( 1, 1 \right)\]
\[x = 0\text{ is the }y - \text{ axis } \]
\[\text{ Consider a horizontal strip of length }= \left| x \right| \text{ and width }= dy\text{ in the first quadrant }\]
\[\text{ Area of approximating rectangle }= \left| x \right| dy\]
\[\text{ Approximating rectangle moves from }y = 1 \text{ to }y = 4 \]
\[\text{ Area of the curve in the first quadrant enclosed by }y = 1\text{ and }y = 4\text{ is the required area of the shaded region }\]
\[ \therefore\text{ Area of the shaded region }= \int_0^4 \left| x \right| dy\]
\[ \Rightarrow A = \int_1^4 x dy ...............\left[ As, x > 0, \left| x \right| = x \right]\]
\[ \Rightarrow A = \int_1^4 \sqrt{\frac{y}{4}} dy \]
\[ \Rightarrow A = \frac{1}{2} \int_1^4 \sqrt{y} dy \]
\[ \Rightarrow A = \frac{1}{2} \left[ \frac{y^\frac{3}{2}}{\frac{3}{2}} \right]_1^4 \]
\[ \Rightarrow A = \frac{1}{2} \times \frac{2}{3}\left[ 4^\frac{3}{2} - 1^\frac{3}{2} \right]\]
\[ \Rightarrow A = \frac{1}{3}\left[ 8 - 1 \right]\]
\[ \Rightarrow A = \frac{7}{3}\text{ sq . units }\]
\[ \therefore\text{ The area enclosed by parabola in the first quadrant and }y = 1, y = 4\text{ is } \frac{7}{3}\text{ sq . units }\]