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When estimating an integral using trapezoidal approximation, we can find the error or uncertainty in the estimation by:
##Error~in~T_n \leq \frac{M(ba)^3}{12n^2}## where ##M## is the maximum value of the absolute value of f''(x) over [a,b], ##n## is the number of intervals, and ##T_n## is the trapezoidal approximation with n intervals.
If we want an estimation within a certain degree of accuracy, we find M, set in less than or equal to our desired maximum value of uncertainty, then solve for n. This will give us an n amount of intervals to use for the approximation to stay withing a specified range of error or uncertainty.
My question is two fold:
First, why is ##M## the absolute value of f''(x)? What is the role that the second derivative plays here and why absolute value?
Second, please see the attached image which is part of an example problem in my text. I don't know what they're doing when solving for the absolute value of f''(x). Clearly, the maximum would be 6e, as when we plug in 1 into the second derivative, this is what we get. But I don't know what they're doing to get this value. I am referring to the very bottom line of the attached image.
##Error~in~T_n \leq \frac{M(ba)^3}{12n^2}## where ##M## is the maximum value of the absolute value of f''(x) over [a,b], ##n## is the number of intervals, and ##T_n## is the trapezoidal approximation with n intervals.
If we want an estimation within a certain degree of accuracy, we find M, set in less than or equal to our desired maximum value of uncertainty, then solve for n. This will give us an n amount of intervals to use for the approximation to stay withing a specified range of error or uncertainty.
My question is two fold:
First, why is ##M## the absolute value of f''(x)? What is the role that the second derivative plays here and why absolute value?
Second, please see the attached image which is part of an example problem in my text. I don't know what they're doing when solving for the absolute value of f''(x). Clearly, the maximum would be 6e, as when we plug in 1 into the second derivative, this is what we get. But I don't know what they're doing to get this value. I am referring to the very bottom line of the attached image.
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