Question:medium
The area bounded by $y = 1 + \dfrac{8}{x^2}$ and the ordinates $x = 2$ and $x = 4$ is
The area bounded by $y = 1 + \dfrac{8}{x^2}$ and the ordinates $x = 2$ and $x = 4$ is
Show Hint
$\displaystyle\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C$. Break compound integrands into simpler terms before integrating.
Updated On: Apr 8, 2026
- 2 sq unit
- 4 sq unit
- $\log 2$ sq unit
- $\log 4$ sq unit
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on Calculus
- Let $f$ be a twice differentiable function such that $f(x) = \int_0^x \tan(t-x) dt - \int_0^x f(t) \tan t dt, x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then $f''(\frac{\pi}{6}) + 12 f'(-\frac{\pi}{6}) + f(\frac{\pi}{6})$ is equal to ________.
- JEE Main - 2026
- Mathematics
- Calculus
- The area of the region {(x, y) : 0 ≤ y ≤ 6 - x, y^2 ≥ 4x - 3, x ≥ 0} is:
- JEE Main - 2026
- Mathematics
- Calculus
- Let $e$ be the base of natural logarithm and let $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$ and $g : \{1, e, e^2, e^3\} \to \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[ f^{-1} \left\{ g^{-1} \left( \frac{1}{2} \right) \right\} \right]^x$, then the area of the region $R = \{ (x, y) : x^2 \le y \le \phi(x), 0 \le x \le 1 \}$ is:
- JEE Main - 2026
- Mathematics
- Calculus
- For the functions $f(\theta) = \alpha \tan^{2}\theta + \beta \cot^{2}\theta$, and $g(\theta) = \alpha \sin^{2}\theta + \beta \cos^{2}\theta$, $\alpha>\beta>0$, let $\min_{0<\theta<\frac{\pi}{2}} f(\theta) = \max_{0<\theta<\pi} g(\theta)$. If the first term of a G.P. is $\left( \frac{\alpha}{2\beta} \right)$, its common ratio is $\left( \frac{2\beta}{\alpha} \right)$ and the sum of its first 10 terms is $\frac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to ________.
- JEE Main - 2026
- Mathematics
- Calculus
- Want to practice more? Try solving extra ecology questions todayView All Questions
