Home > Calculus > Definite and indefinite integrals

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

Answer 1

Christopher Allers

#int_0^pi(1+sinx)dx = int_0^pi dx +int_0^pi sinx dx = (x-cosx)|_0^pi = pi#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

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).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

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.