# How do you evaluate the definite integral #int 1+sinx # from #[0,pi]#?

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To evaluate the definite integral (\int_{0}^{\pi} (1 + \sin(x)) , dx), you can follow these steps:

- Integrate the function with respect to ( x ):

[ \int (1 + \sin(x)) , dx ]

[ = \int 1 , dx + \int \sin(x) , dx ]

[ = x - \cos(x) + C ]

- Evaluate the definite integral by substituting the upper and lower limits of integration:

[ \int_{0}^{\pi} (1 + \sin(x)) , dx = [\pi - \cos(\pi)] - [0 - \cos(0)] ]

[ = [\pi + 1] - [0 + 1] ]

[ = \pi ]

Therefore, the value of the definite integral (\int_{0}^{\pi} (1 + \sin(x)) , dx) is (\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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