How would you use integration to find the area of the triangle with vertices (0,0), (2,2) and (4,0)?
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The line segment from (0,0) to (2,2) has equation y = x.
The line segment from (2,2) to (4,0) has equation y = 4 - x.
The line segment from (0,0) to (4,0) lies on the x-axis.
So the area of the triangle is
integral 0 to 2 of x dx + integral 2 to 4 of (4-x) dx
= [(1/2)x^2 from 0 to 2] + [(4x - (1/2)x^2) from 2 to 4]
= [(1/2)(2^2) - 0] + [4(4) - (1/2)(4^2) - (4(2) - (1/2)(2^2))]
= (2 - 0) + (8 - 6)
= 4.
Of course, 4 is the expected answer because A = (1/2)bh = (1/2)(4)(2) = 4.
Lord bless you today!
The line segment from (2,2) to (4,0) has equation y = 4 - x.
The line segment from (0,0) to (4,0) lies on the x-axis.
So the area of the triangle is
integral 0 to 2 of x dx + integral 2 to 4 of (4-x) dx
= [(1/2)x^2 from 0 to 2] + [(4x - (1/2)x^2) from 2 to 4]
= [(1/2)(2^2) - 0] + [4(4) - (1/2)(4^2) - (4(2) - (1/2)(2^2))]
= (2 - 0) + (8 - 6)
= 4.
Of course, 4 is the expected answer because A = (1/2)bh = (1/2)(4)(2) = 4.
Lord bless you today!
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between x=0 and x=2, the equation of the top of the triangle is y=x
between x=2 and x=4, the equation of the top of the triangle is y = -x + 4
The elemental area of a width dx of height y is ydx
So the area is{ S1= integral of ydx =xdx from x=0 to x=2.} plus the integral S2 of ydx= (-x +4)dx from x=2 to x=4
The answer is A=S1 + S2 = 4
between x=2 and x=4, the equation of the top of the triangle is y = -x + 4
The elemental area of a width dx of height y is ydx
So the area is{ S1= integral of ydx =xdx from x=0 to x=2.} plus the integral S2 of ydx= (-x +4)dx from x=2 to x=4
The answer is A=S1 + S2 = 4