How do you find the area between y=−38x(x−8),y=10−12x,x=2,x=8?

2 Answers
James
Apr 29, 2018

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Explanation:

y1=−38x(x−8)

y2=10−12x

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this is a sketch for your functions you can use this website to sketch them[www.desmos.com]

the area between the two curve equal

A=∫82(y2−y1)⋅dx

A=∫82(10−12x)−(−38x(x−8))⋅dx

A=∫82(10−12x)+(38x(x−8))⋅dx

A=∫82(10−12x)⋅dx+∫82(38x(x−8))⋅dx

∫(10−12x)⋅dx=10⋅x−ln(|x|)2

∫82(10−12x)⋅dx=ln(2)−402−ln(8)−1602=−ln(8)−ln(2)−1202=59.31

∫(38x(x−8))=−3⋅(ln(|x|)−ln(|x−8|))64=−3⎛⎜⎝ln(|x|)−(((lnxln8)))64⎞⎟⎠

∫82(38x(x−8))=

then complete the steps normally

Mark D.
Apr 30, 2018

18units2

Explanation:

x=2 and x=8 are two vertical lines and y=10−12x is a diagonal line that passes through x=2 at (2,9) and x=8 at (8,6). This gives us a trapezium that has two parallel sides of length 9 and 6 and a height of 6 so the area would be

A=12(9+6)6=45units2

y=−38x(x−8) ⇒y=−38x2+3x

This is an ∩ shaped parabola that passes through x=2andx=8 . If we find the area under the curve by intergration

∫82 −38x2+3x dx

[−18x3+32x2]82

[−5128+1922]−[−88+122]

32−5=27

So the area enclosed by all four graphs is the area of the trapezium less the area under the curve.

45 - 27 = 18

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