On the interval left[ frac{5pi}{3}, frac{7pi} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$.By the Fundamental Theorem of Calculus,$f'(x) = 6\cos x - 2\sin x$.In the interval $\left

Vedclass · Accepted Answer

$3\sqrt{3} - 2\sqrt{2} - 1$

Vedclass · Answer

(B) Given $f(x) = \int_{5\pi/3}^x (6\cos t - 2\sin t) \, dt$.By the Fundamental Theorem of Calculus,$f'(x) = 6\cos x - 2\sin x$.In the interval $\left[ \frac{5\pi}{3}, \frac{7\pi}{4} \right]$,$\cos x > 0$ and $\sin x  0$.Since $f'(x) > 0$,$f(x)$ is an increasing function on the given interval.Therefore,the greatest value occurs at the upper limit $x = \frac{7\pi}{4}$.$f\left( \frac{7\pi}{4} \right) = \int_{5\pi/3}^{7\pi/4} (6\cos t - 2\sin t) \, dt = [6\sin t + 2\cos t]_{5\pi/3}^{7\pi/4}$.$= \left( 6\sin\frac{7\pi}{4} + 2\cos\frac{7\pi}{4} \right) - \left( 6\sin\frac{5\pi}{3} + 2\cos\frac{5\pi}{3} \right)$.$= \left( 6\left(-\frac{1}{\sqrt{2}}\right) + 2\left(\frac{1}{\sqrt{2}}\right) \right) - \left( 6\left(-\frac{\sqrt{3}}{2}\right) + 2\left(\frac{1}{2}\right) \right)$.$= \left( -\frac{6}{\sqrt{2}} + \frac{2}{\sqrt{2}} \right) - (-3\sqrt{3} + 1)$.$= -\frac{4}{\sqrt{2}} + 3\sqrt{3} - 1 = 3\sqrt{3} - 2\sqrt{2} - 1$.

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:

Similar Questions

$\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x$ is equal to

If $f(x) = |x - 1|$,then $\int_0^2 {f(x)dx}$ is

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$,then the value of the integral $\int_{0}^{\theta} \sin x \,dx$ is equal to:

If $m$ and $n$ are integers such that $(m - n)$ is an odd integer,then the value of the definite integral $\int_{0}^{\pi} \cos(mx) \sin(nx) \, dx$ is:

Prove that $\int_{0}^{\frac{\pi}{2}} \sin^{3} x \, dx = \frac{2}{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module