If intlimits_e^x {t,f(t),dt = sin x - xcos x | vedclass

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

Question

(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.Applying the Leibniz integral rule,we differentiate both

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.Applying the Leibniz integral rule,we differentiate both sides with respect to $x$:$\frac{d}{{dx}}\left[ {\int\limits_e^x {t\,f(t)\,dt} } ight] = \frac{d}{{dx}}\left[ {\sin x - x\cos x - \frac{{{x^2}}}{2}} ight]$Using the product rule on the right side:$x\,f(x) = \cos x - (\cos x - x\sin x) - x$$x\,f(x) = \cos x - \cos x + x\sin x - x$$x\,f(x) = x\sin x - x$Dividing by $x$ (since $x 
eq 0$):$f(x) = \sin x - 1$Now,substitute $x = \frac{\pi}{6}$:$f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) - 1$$f(\frac{\pi}{6}) = \frac{1}{2} - 1 = -\frac{1}{2}$.

If $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ for all $x \in R - \{0\}$,then the value of $f(\frac{\pi}{6})$ is

Similar Questions

$\int_{-5 \pi}^{5 \pi} (1-\cos 2x)^{\frac{5}{2}} dx =$

The value of the integral $\int_0^{\pi / 2} \sin^5 x \, dx$ is

$\int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx =$

If $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$,then the value of $f'(\frac{\pi}{2})$ is:

$\int_0^{\pi /2} \sin^5 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module