# What is the net area between #f(x) = sqrt(x^2+2x+1) # and the x-axis over #x in [2, 4 ]#?

Actually we can do this one without the need of Calculus, as follows.

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To find the net area between ( f(x) = \sqrt{x^2+2x+1} ) and the x-axis over the interval ([2, 4]), we need to compute the definite integral of ( f(x) ) from ( x = 2 ) to ( x = 4 ) and take the absolute value of the result.

The definite integral of ( f(x) = \sqrt{x^2+2x+1} ) from ( x = 2 ) to ( x = 4 ) is:

[ \int_{2}^{4} \sqrt{x^2+2x+1} , dx ]

This integral can be solved by substitution or by recognizing that ( \sqrt{x^2+2x+1} ) is equivalent to ( |x + 1| ).

Evaluating the integral:

[ \int_{2}^{4} \sqrt{x^2+2x+1} , dx = \int_{2}^{4} |x + 1| , dx ]

[ = \left[ \frac{1}{2}(x+1)^2 \right]_{2}^{4} ]

[ = \left( \frac{1}{2}(4+1)^2 \right) - \left( \frac{1}{2}(2+1)^2 \right) ]

[ = \left( \frac{1}{2}(5)^2 \right) - \left( \frac{1}{2}(3)^2 \right) ]

[ = \left( \frac{1}{2}(25) \right) - \left( \frac{1}{2}(9) \right) ]

[ = \frac{25}{2} - \frac{9}{2} ]

[ = \frac{16}{2} ]

[ = 8 ]

Taking the absolute value, the net area between ( f(x) = \sqrt{x^2+2x+1} ) and the x-axis over the interval ([2, 4]) is ( 8 ).

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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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