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What is the net area between #f(x)=x^3+4x^2-2# in #x in[2,5] # and the x-axis?

Answer 1

Scarlett Allison

#area=260.58 " unit"#

#area=int_2^5 f(x) d x#

#area=int _2^5 (x^3+4x^2-2) d x#

#area=|1/4 x^4+4*1/3 x^3-2x|_2^5#

#area=|1/4 x^4+4/3x^3-2x|_2^5#

#area=(1/4 * 5^4+4/3 * 5^3-2*5)-(1/4 * 2^4+4/3 * 2^3-2*2)#

#area=(625/4+500/3-10)-(4+32/3-4)#

#area=625/4+500/4-10-4-32/3+4#

#area=1125/4-10-32/3#

#area=(3375-120-128)/12#

#area=260.58 " unit"#

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Answer 2

Charles Arocha

To find the net area between the function ( f(x) = x^3 + 4x^2 - 2 ) and the x-axis in the interval ([2,5]), you need to integrate the absolute value of the function from ( x = 2 ) to ( x = 5 ). Mathematically, this is represented as:

[ \text{Net Area} = \int_{2}^{5} |f(x)| , dx ]

First, determine the intervals where the function is above and below the x-axis in the interval ([2,5]). Then integrate the function over those intervals, taking the absolute value to ensure a positive area.

[ \text{Net Area} = \int_{2}^{3} (x^3 + 4x^2 - 2) , dx + \int_{3}^{5} -(x^3 + 4x^2 - 2) , dx ]

Calculate each integral separately and add their absolute values to find the net area.

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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.