Find the approximation?

Question

If the integral of (x^2-2x+2)dx from 0 to 6 is approximated by a lower sum using three inscribed rectangles of equal width on the x-axis, find the approximation. 

Answer 1

"If the integral of (x^2-2x+2)dx from 0 to 6 is approximated by a lower sum using three inscribed rectangles of equal width on the x-axis, find the approximation."

f(x) = x^2 - 2x + 2 = (x - 1)^2 + 1,
which is a parabola shifted right 1 and up 1.

The bases of each rectangle will be 6/3 = 2 units.

The heights of the rectangles will be the smallest f(x) on the closed intervals [0,2],[2,4],[4,6].

On [0,2] the smallest f(x) is at the vertex, f(1) = 1.

On [2,4] the smallest f(x) is f(2) = 2.

On [4,6] the smallest f(x) is f(4) = 10.

So the Lower Sum = 2(1+2+10) = 26 units^2

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Bonus: Upper Sum = 2(2+10+26) = 76

Avg = 50

Answer 2

∫^{6 ( X^2 - 2X +2) dx = X^3/3  - X^2 + 2X + C}

₀

                             = 6^3/ 3 - (6 ^2) + 2( 6 ) - ( 0 ) =

 

                              = 72 - 36 +12 = 48