# How do you evaluate the definite integral #int sinxdx# from #[pi/3, pi]#?

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To evaluate the definite integral (\int_{\frac{\pi}{3}}^{\pi} \sin x , dx), you can use the fundamental theorem of calculus and the properties of the sine function. First, find the antiderivative of (\sin x), which is (-\cos x). Then, evaluate (-\cos x) at the upper limit ((\pi)) and subtract its value at the lower limit ((\frac{\pi}{3})).

(\int_{\frac{\pi}{3}}^{\pi} \sin x , dx = [-\cos x]_{\frac{\pi}{3}}^{\pi} = -\cos(\pi) - (-\cos(\frac{\pi}{3})) = -(-1) - \left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2})

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To evaluate the definite integral ( \int_{\pi/3}^{\pi} \sin(x) , dx ), we can use the properties of the sine function and definite integrals. We integrate the sine function with respect to ( x ) over the interval ([ \pi/3, \pi ]), and then subtract the value of the integral at the lower limit from the value of the integral at the upper limit. This gives us the area under the curve of ( \sin(x) ) between ( \pi/3 ) and ( \pi ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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