If f(x) = int_0^x {t(sin x - sin t) dt},then | vedclass

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(A) Given $f(x) = \int_0^x t(\sin x - \sin t) dt = \sin x \int_0^x t dt - \int_0^x t \sin t dt$.Evaluating the integrals:$\int_0^x t dt = \frac{x^2}{2

Vedclass · Accepted Answer

$f'''(x) + f'(x) = \cos x - 2x \sin x$

Vedclass · Answer

(A) Given $f(x) = \int_0^x t(\sin x - \sin t) dt = \sin x \int_0^x t dt - \int_0^x t \sin t dt$.Evaluating the integrals:$\int_0^x t dt = \frac{x^2}{2}$.Using integration by parts for $\int t \sin t dt = -t \cos t + \sin t$.So,$f(x) = \sin x (\frac{x^2}{2}) - [-t \cos t + \sin t]_0^x = \frac{x^2}{2} \sin x + x \cos x - \sin x$.Now,find the derivatives:$f'(x) = x \sin x + \frac{x^2}{2} \cos x + \cos x - x \sin x - \cos x = \frac{x^2}{2} \cos x$.$f''(x) = x \cos x - \frac{x^2}{2} \sin x$.$f'''(x) = \cos x - x \sin x - x \sin x - \frac{x^2}{2} \cos x = \cos x - 2x \sin x - \frac{x^2}{2} \cos x$.Check $f'''(x) + f'(x)$:$f'''(x) + f'(x) = (\cos x - 2x \sin x - \frac{x^2}{2} \cos x) + (\frac{x^2}{2} \cos x) = \cos x - 2x \sin x$.Thus,option $A$ is correct.

If $f(x) = \int_0^x {t(\sin x - \sin t) dt}$,then which of the following is true?

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