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Area under curve (integration) ?
y=(2x-2)(x-2)(x-6) determine the area between the x axis from x = 1 to x = 6...I expanded the brackets and got 5x^2-2x+28 and used integration rule and got 5/3(x^3) - 14x^2 + 28x...is this right? and what is the answer...this is my lecturers answer : 113.5 units^2 i think its incorrect.
1 Answer
- Fred OsimLv 59 years agoFavorite Answer
y=(2x-2)(x-2)(x-6)=(2x-2)(x^2-8x+12)
=2x^3-16x^2+24x-2x^2+16x-24
=2x^3-18x^2+40x-24
now area A=sum|for x=1 to 6|
=(2(x^4)/4-18(x^3)/3+40(x^2)/2-24x)|for x=1 to 6
so intgl y=(x^4)/2-6x^3+20x^2-24x|for x=1 to 6
we know y=(2x-2)(x-2)(x-6)...and now find the zero...
so x=1 or x=2 or x=6..
so from x=1 to 2, the curve must be above the x axis.
so A={2^4/2-6(2^3)+20(2^2)-24(2)}
-{1/2-6+20-24}
=-8+9.5=1.5
so from x=2 to 6, the curve must be below the x axis
so A=(6^4)/2-6(6^3)+20(6^2)-24(6)+8
so A=-72+8=-64
total area=|-64|+|1.5|=65.5.......ans
somehow, 72+9.5+8+8+8+8=113.5
don't know where i went wrong....pls check calculation.
