Question

A question related to integrals.

Answer 1

@zepdrix

Answer 2

Taking a look at your function, it is positive. Since the function is positive, its integral is also positive and so we get\[area > 0 \tag{1}\]

The maximum value of \(e^x(x-1)^4\) in \([0, 1]\) can be calculated as follows:\[f'(x) = 4e^x (x-1)^3 + e^x(x-1)^4 = (x-1)^3 \cdot \left(4e^x + x-1\right)\]If you see carefully, \(f'(x) \) is negative in the interval \([0,1]\) (a decreasing function). This means that the maximum value can only be at \(x=0\) and that maximum value is \(f(0) = 1\).\[f(x) \le 1\]\[\int_0^1 f(x) dx \le \int_0 1 dx = 1\]And so,\[area < 1\tag{2}\]

Answer 3

Two questions
1. How did you made \( \le \text{into} <\)?
2. Isn't integral the SUM of areas of all bars from x=0 to x=1? If yes, then if the maximum occurs at x=0 and the area at that point is equal or a little less than 1, wouldn't the areas of other bars (when added) make the whole area greater than 1?

Answer 4

@FaiqRaees
I proceed differently.
The question says:
"considering the area bounded by....",
So I use the exact area as a starting point, equal to 24e-65 (not\(24e^{-65}\)!).
The two factors in the integral are:
(1-x)^4, which is a strictly decreasing function from 1 to zero, and
e^x, which is a strictly increasing function from 1 to e.

The two approximation are respectively taking a chord between the end points of the factors.
(A)
From the graph of e^x, we know that the area is greater than 1 (=e^0), so
24e-65>1 or e>65/24.
(B)
Considering both graphs, if we make the e^x graph a trapezoid by joining the end points, the integral of the product would be greater than A=24e-65.
The integral of a quartic spandrel (0 to M) and a trapezoid (N1,N2) over an inteval of L is known to be:
L(M)(N1+5N2)/30 > area
Substituting
L=1, M=1, N1=1, N2=e
1(1)(1e+5(1))/30 > 24e-65
e+5 > 720e-1950
719e < 1955
e < 1955/719 =2.719 (which is tighter than 11/4=2.75)
If we had used a straight line, then the integral would be
L(M)(N1+3N2)/12=1(1)(e+3)/12 > 24e-65
e+3>144e-390
e<393/143=2.748 (closer to 11/4 !)