Exams › JEE Main › Maths

Find the area enclosed by the graph of y = ln(x) and the straight lines y = 0, y = ln(3), and x = 0.

1. 3
2. 3 ln(3) - 2
3. 3 ln(3) + 2
4. 2

Correct answer: 2

Solution

The region is bounded left by x=0 and right by x=e^y, for y from 0 to ln3. Area = integral_0^(ln3) e^y dy = e^(ln3)-1 = 3-1 = 2.

Related JEE Main Maths questions

• For the function y = a²/x + b²/(a - x), where a > 0 and b > 0, the least value of y for x in the interval (0, a) is:
• Let f(x)=∫ from 2x to sin x of cos(t³) dt. Then the value of f''(x) is
• Evaluate the integral ∫_(√(ln 3))^(√(ln 2)) (xsin(x²))/(sin(x²)+sin(ln 6-x²)) dx.
• Evaluate the integral ∫ (e^x (1+xⁿ⁻¹-xⁿ))/((1-xⁿ)√(1-x²ⁿ)) dx.
• Let J = ∫ from 0 to 1 of (cos x)/√x dx and K = ∫ from 0 to 1 of (sin x)/√x dx. Which of the following statements is correct?
• A curve passes through the point (0,1) and satisfies y = ∫ x² e^(x³) dx. If this curve is expressed as x = f(y), then f(y) equals

⚔️ Practice JEE Main Maths free + battle 1v1 →