The area (in sq. units) bounded by the parabola \( y^2=x \) and line \( x=4 \) in first quadrant is equal to \( A \), then which of the following statements is/are TRUE ?
A. \( A=\displaystyle\frac{16}{3} \)
B. \( A=\int_0^4\displaystyle \sqrt{x}\,dx \)
C. \( A=2\int_0^4\displaystyle \sqrt{x}\,dx \)
D. \( A=\int_0^2\displaystyle y^2\,dy \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A and B only
In the first quadrant the parabola is \( y=\sqrt{x} \), and the region lies under this curve from x = 0 to x = 4.
\( \displaystyle A=\int_0^4\sqrt{x}\,dx \)
\( \displaystyle =\left[\frac{2}{3}x^{3/2}\right]_0^4 \)
\( \displaystyle =\frac{2}{3}\times 8=\frac{16}{3} \)
So \( A=\int_0^4\sqrt{x}\,dx=\frac{16}{3} \).
Checking the statements
(A) \( A=\displaystyle\frac{16}{3} \) — Correct. The area is \( \frac{16}{3} \).
(B) \( A=\int_0^4\displaystyle\sqrt{x}\,dx \) — Correct. This is the area under \( y=\sqrt{x} \) from x = 0 to x = 4.
(C) \( A=2\int_0^4\displaystyle\sqrt{x}\,dx \) — Incorrect. This is \( \frac{32}{3} \), which also adds the part below the x-axis.
(D) \( A=\int_0^2\displaystyle y^2\,dy \) — Incorrect. This is \( \frac{8}{3} \), the area between the parabola and the y-axis.