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)# ?

Question

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)# ?

Paisley Johnson · Accepted Answer

#A(x) = 11x^2 + 8x#
#A'(x) = 22x+8 = f(x)# The area under #f(x) = 22x+8# and the x-axis #in (0,x)# consists of a triangle of base #x# and height #f(x)-8# plus a rectangle of sides #8# and #x#. Area of triangle = #1/2 xx Base xx Height# Area of rectangle = #Base xx Height# Therefore in this case: Area of triangle #= 1/2x(22x+8-8)# = #11x^2# Area of rectangle #=8x# If the required area is denoted by #A(x)# #A(x) = 11x^2+8x# Also #A'(x) = d/dx(11x^2+8x) = 22x+8 = f(x)# Therefore we have confirmed that #A'(x) = f(x)#

Cameron Alexander · Answer

undefined 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) $.