# Given #f(x) = 22x+8#. Then let the area bounded by #f(x): x in (0,x)# and the x-axis be denoted by #A(x)#. Show that #A'(x) = f(x)# ?

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To find ( A'(x) ), we need to differentiate the expression for the area ( A(x) ) with respect to ( x ).

Given ( f(x) = 22x + 8 ), the expression for the area ( A(x) ) bounded by ( f(x) ) and the x-axis from 0 to ( x ) is given by the integral:

[ A(x) = \int_{0}^{x} f(t) , dt ]

Substituting ( f(t) = 22t + 8 ) into the integral, we get:

[ A(x) = \int_{0}^{x} (22t + 8) , dt ]

Now, we differentiate ( A(x) ) with respect to ( x ) using the Fundamental Theorem of Calculus:

[ A'(x) = \frac{d}{dx} \left( \int_{0}^{x} (22t + 8) , dt \right) ]

[ A'(x) = 22x + 8 ]

Hence, ( A'(x) = f(x) ).

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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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