The area of the region lying in the first quadrant and bounded by \( y=8x^2,x=0,y=1 \) and \( y=4 \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{7}{3\sqrt{2}} \)
The region lies between the curve and the y-axis, from y = 1 to y = 4, so the area is found by integrating with respect to y. In the first quadrant:
\( \displaystyle y=8x^2\ \Rightarrow\ x=\sqrt{\frac{y}{8}} \)
\( \displaystyle =\frac{\sqrt{y}}{2\sqrt{2}} \)
\( \displaystyle \text{Area}=\int_{1}^{4}\frac{\sqrt{y}}{2\sqrt{2}}\,dy \)
\( \displaystyle =\frac{1}{2\sqrt{2}}\cdot\frac{2}{3}\left[y^{3/2}\right]_{1}^{4} \)
\( \displaystyle =\frac{1}{3\sqrt{2}}(8-1) \)
\( \displaystyle =\frac{7}{3\sqrt{2}} \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{7}{\sqrt{2}} \) — Incorrect. The area is \( \frac{7}{3\sqrt{2}} \), not \( \frac{7}{\sqrt{2}} \).
Option (3) \( \displaystyle\frac{14}{3\sqrt{2}} \) — Incorrect. This is twice the area: it counts the region on both sides of the y-axis, but only the first quadrant is required.
Option (4) \( \displaystyle\frac{7}{3} \) — Incorrect. The area is \( \frac{7}{3\sqrt{2}} \), not \( \frac{7}{3} \).