25: The derivative

is not simple enough to be handled easily. The standard procedure (finding its crucial points, that is, zero points od the next, the fourth derivative) leads to an equation of degree 3 that takes time to solve, so it is much easier (though we will loose a bit) to first make the estimate a bit more generous by taking maximum over the whole interval [0,1] rather then just over [0,x] for some x ≤ 1. Then we observe that (12t−8t³) is positive and increasing on [0,1], so its maximum is 4 there (the value at the right endpoint), while the exponential is positive and decreasing on [0,1] and hence its maximum is 1 there (its value at its left endpoint). Putting it together we see that the desired maximum is surely at most 4. Thus

Now we use it in estimating the error of integration.

Answer