If f is a continuous function and f(x+T)=f(x) | vedclass

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(B) Given the integral $I = \int_0^{50\pi} \sqrt{1-\cos 2x} dx$. Using the identity $1-\cos 2x = 2\sin^2 x$,we have $\sqrt{1-\cos 2x} = \sqrt{2\sin^2

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$100\sqrt{2}$

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(B) Given the integral $I = \int_0^{50\pi} \sqrt{1-\cos 2x} dx$. Using the identity $1-\cos 2x = 2\sin^2 x$,we have $\sqrt{1-\cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}|\sin x|$. Since the function $f(x) = \sqrt{2}|\sin x|$ has a period $T = \pi$,we can use the property $\int_0^{NT} f(x) dx = N \int_0^T f(x) dx$. Here $N = 50$ and $T = \pi$,so $I = 50 \int_0^{\pi} \sqrt{2}|\sin x| dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$. Thus,$I = 50\sqrt{2} \int_0^{\pi} \sin x dx$. $I = 50\sqrt{2} [-\cos x]_0^{\pi}$. $I = 50\sqrt{2} (-(\cos \pi - \cos 0)) = 50\sqrt{2} (-(-1 - 1)) = 50\sqrt{2} (2) = 100\sqrt{2}$.

If $f$ is a continuous function and $f(x+T)=f(x)$ for all $x \in R$,it is given that $\int_0^{NT} f(t) dt = N \int_0^T f(t) dt$ (where $N$ is a natural number). Then,evaluate $\int_0^{50\pi} \sqrt{1-\cos 2x} dx$.

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