Answer:
The area of the region between the two curves by integration over the x-axis is 9.9 square units.
Step-by-step explanation:
This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

Given that resulting expression is a second order polynomial of the form
, there are two real and distinct solutions. Roots of the expression are:
and
.
Now, it is also required to determine which part of the interval
is equal to a number greater than zero (positive). That is:


and
.
Therefore, exists two sub-intervals:
and
. Besides,
in each sub-interval. The definite integral of the region between the two curves over the x-axis is:




The area of the region between the two curves by integration over the x-axis is 9.9 square units.
Answer:
The value of x that gives the maximum transmission is 1/√e ≅0.607
Step-by-step explanation:
Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

We need to equalize f' to 0
- k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
- 2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
- 2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
- ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
- 1/x = e^(1/2)
- x = 1/e^(1/2) = 1/√e ≅ 0.607
Thus, the value of x that gives the maximum transmission is 1/√e.
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Answer:
Tn=13n-18
Step-by-step explanation:
Tn=-5(n-1)13
Simplify
this is using the a+(n-1)d