Integral Calculus
Area bisection by ordinate; definite integral
MMTS_Full_Test_10
Grade 12
Question:
The area bounded by the $x$-axis, part of the curve $y=1+x^{-2}$ and the ordinates $x=1$, $x=2$ is divided into equal parts by the ordinate at $x=a$ such that $a=\dfrac{3+\sqrt{p}}{8}$. The value of $p$ is
(A) 70
(B) 71
(C) 72
(D) 73
Step-by-Step Solution
Key Concept: Total area $=\int_1^2(1+x^{-2})dx=1+1/2=3/2$. Set $\int_1^a(1+x^{-2})dx=3/4$: $(a-1-1/a+1)=3/4\Rightarrow a-1/a=3/4$. Solve: $4a^2-3a-4=0$.
$p=73$.
Correct Answer: (D) 73