Question:
The area enclosed between the curve \(y = \log_e(x + e)\) and the coordinate axes is:
The area enclosed between the curve \(y = \log_e(x + e)\) and the coordinate axes is:
Show Hint
Use integration by parts to calculate the area under logarithmic curves.
Updated On: Sep 16, 2025
- 1
- 2
- 3
- 4
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
The area under the curve is given by: \[ A = \int_0^\infty \log_e(x + e) \, dx \] Use integration by parts or known results to evaluate: \[ A = 2 \] Thus, the area enclosed is \(2\).
Was this answer helpful?
0
0
Top Questions on Fundamental Theorem of Calculus
- If log$_{10}$(x + 1) = 2, what is the value of x?
- BITSAT - 2025
- Mathematics
- Fundamental Theorem of Calculus
- Find the derivative of \( y = \sin(x^2) \) with respect to \( x \).
- BITSAT - 2025
- Mathematics
- Fundamental Theorem of Calculus
- The area enclosed between the curve \(y = \log_e(x + e)\) and the coordinate axes is:
- BITSAT - 2025
- Mathematics
- Fundamental Theorem of Calculus
- Evaluate: $$ \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} $$
- BITSAT - 2025
- Mathematics
- Fundamental Theorem of Calculus
- Evaluate the sum: $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} $$
- BITSAT - 2025
- Mathematics
- Fundamental Theorem of Calculus
View More Questions
Questions Asked in BITSAT exam
- A particle moves along the x-axis under a force $ F(x) = 6x^2 $ N. The work done by this force in moving the particle from $ x = 1 \, \text{m} $ to $ x = 2 \, \text{m} $ is:
- BITSAT - 2025
- work, energy and power
- A body starts from rest and moves with constant acceleration. If it covers a distance of $ 40 \, \text{m} $ in 4 seconds, then its acceleration is:
- BITSAT - 2025
- Kinematics
- A simple pendulum has a time period of 2 s on Earth's surface. What is its time period at a height equal to the Earth's radius (R)? (Acceleration due to gravity at height h is \( g_h = \frac{g}{(1 + h/R)^2} \)).
- BITSAT - 2025
- Mechanics
- A projectile is launched with an initial velocity of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) with the horizontal. What is the maximum height reached by the projectile? (Take \( g = 10 \, \text{m/s}^2 \))
- BITSAT - 2025
- Projectile motion
- If $ x + \frac{1}{x} = 2 \cos \theta $, find the value of \[ x^5 + \frac{1}{x^5} \] in terms of $\theta$.
- BITSAT - 2025
- Algebra
View More Questions