Calculus problem, part c only?

The solution is 4(1 + 4 + 6) = 44. But I just can't understand the intervals used to get these numbers. delta x would be 4, but I don't know what to do after that. Do I divide the intervals according to m(x)'s values? Sorry, any help and explanations would be appreciated. This is just for part c.

https://s23.postimg.org/6ycqrm8x7/Screen_Shot_2017...

Comments

  •  

    m⁻¹(x) is the inverse of m(x)

    So take the table for m(x) and switch the 2 rows, and label them x and m⁻¹(x)

    i.e. the row that was m(x) is now at top and labeled x

    and the row that was x is now at bottom and labeled m⁻¹(x)

    x      0  2  3  4  6  9  10  11  12

    m⁻¹(x)    0  1  2  3  4  5  6  7  8

    We are integrating m⁻¹(x) over the interval [0, 12] with 3 sub-intervals:

    [0, 4], [4, 8], and [8,12] ----> each sub-interval has width 4

    and using midpoints: x = 2, 6, 10

    ∫ [0 to 12] m⁻¹(x) dx

    = 4 * m⁻¹(2) + 4 * m⁻¹(6) + 4 * m⁻¹(10)

    = 4 (m⁻¹(2) + m⁻¹(6) + m⁻¹(10))

    = 4 (1 + 4 + 6)

    = 44

  • You're integrating m^-1, so the y axis is actually the "x" row. Heights of rectangles will be those values. The x axis is the "m(x)" row.

    Perhaps it would help to write g(x) next to the first row and x next to the second row.

    You are correct, the rectangles are width 4, going from 0-4, 4-8, 8-12. The midpoints are at 2, 6, 10. The heights according to the midpoint rule aré the "function" value (1st row) when the 2nd row is 2, 6, 10. Those values are 1, 4, and 6.

    So the sum of the three areas is 4*1 + 4*4 + 4*6.

Sign In or Register to comment.