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How do you evaluate the definite integral #int (2+x)dx# from [0,4]?

Answer 1

William Abdalla

16

#int_0^4(2+x)dx#

applying the rule of power

#intax^ndx=ax^(n+1)/(n+1)+C; n!=-1#

we have

#int_0^4(2+x)dx=[2x+x^2/2]_0^4#

There's no need for the limited constant.

#[2x+x^2/2]^4-cancel([2x+x^2/2]_0)#

#(2xx4+4^2/2)=8+8=16#

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Answer 2

Emma Aldridge

To evaluate the definite integral (\int_{0}^{4} (2+x) , dx), you can follow these steps:

Find the antiderivative of the integrand (2+x) with respect to (x), which is (2x + \frac{1}{2}x^2).
Evaluate the antiderivative at the upper limit of integration (4) and subtract the value of the antiderivative at the lower limit of integration (0).

[ \int_{0}^{4} (2+x) , dx = \left[2x + \frac{1}{2}x^2\right]_{0}^{4} = \left(2(4) + \frac{1}{2}(4)^2\right) - \left(2(0) + \frac{1}{2}(0)^2\right) ] [ = (8 + 8) - (0 + 0) = 16 ]

Therefore, the value of the definite integral is 16.

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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.