Area Under the Curve
Difference of Two Bounded Areas in First Quadrant
nta_pyq_2026_jan
Grade 12
Question:
Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and $y$-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$, and $y$-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to
$\dfrac{2}{3}(3\sqrt{2}+1)$
$\dfrac{2}{3}(2\sqrt{2}+1)$
$\dfrac{2}{3}(\sqrt{2}+1)$
$\dfrac{2}{3}(4\sqrt{2}+1)$
Step-by-Step Solution
Key Concept: $A_1$: intersection of $y=x^2+2$ and $x+y=8$ at $x=2,y=6$. $A_1=\int_0^2(8-x-(x^2+2))dx=\int_0^2(6-x-x^2)dx=\frac{22}{3}$. $A_2$: bounded by $y=x^2+2$, $y^2=x$ (i.e., $x=y^2$), $x=2$, $y$-axis.
$A_1=22/3$, $A_2=4\sqrt{2}/3$. $A_1-A_2=\frac{2}{3}(2\sqrt{2}+1)$.
Correct Answer: 2