On the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,the greatest value of the function $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$ is:

  • A
    $3\sqrt{3} + 2\sqrt{2} + 1$
  • B
    $3\sqrt{3} - 2\sqrt{2} - 1$
  • C
    Does not exist
  • D
    None of these

Explore More

Similar Questions

$\int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx = $

$\int_{0}^{\frac{\pi}{2}} \frac{4x \sin x + x^2 \cos x}{2\sqrt{\sin x}} dx$ is equal to

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

The value of $\int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$ is

$\int_{-2}^{1} [x+1] \, dx =$ (Where $[x]$ is the greatest integer function not greater than $x$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo