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Calculus/Integral of absolute value function

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Expert: Alon Mandes - 5/31/2009


Question
I need some help with this problem:

Evaluate the integral of the absolute value of (2x-3) from x=8 to x=0.

Thanks!

Answer
The function f(x)=|2x-3| can be defined also as :
f(x)=-2x+3 when x<3/2 & 2x-3 when x>3/2. Thus our integral will be:
8           1.5            8
∫|2x-3| dx = ∫(-2x+3) dx + ∫ (2x-3) dx
2           0             1.5

=-x²+3x {0->1.5} + x²-3x {1,5->8}
=9/4 + 9/2 + 64 - 24 - 9/4 + 9/2
=49 .

Alon.

Calculus

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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M.A in Mathematics & Bs.c in Electronics.

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