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Evaluate $ \displaystyle \int^2_{-2} (x + 3) \sqrt{4 - x^2} \,dx $ by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.

6$\pi$

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We have a question in which we need to evaluate -2 is uh, to to lower limited -2 and a parliament is too. So this is a definite integral With limit -2 x plus three. Four minus x squared. Underwrote dx. We need to express this as some of to interiors. A lot of hrodriguez Limit -2X. Just multiplying with this for minus X squared dx blessed three minus two 22, 4 -1. Esquire dx mm No, since this is an odd function and if this is an art function and limited symmetrical so integral will be equal to zero. But this will depend as it is for minor sex esquire dx. Now four minus x square is the upper half of the circle. Access squared plus one is equal to four. This is the upper half of their Circle, this is -2. So this is the area of the area out there. Uh, upper half of the circle. This is the area of the upper half of the circle. Okay, so simply will be just writing it will be using geometry tool solve this radius is definitely too. So this will be by our squad by to where radius is too. So three bye bye to into two squares. So 6565 squired unit should be the answer. Thank you

Chandigarh University