Answer:
(A) true.
(B). False.
Step-by-step explanation:
(A). F(x) = 4xe^-2x.
Let's make the assumption that 2x = b -----------------------(1).
Therefore, taking the differentiation with respect to x, we have;
2x dx = db.
The next thing thing to do is to integrate, taking the upper limit to be infinity and the lower limit to be zero:
∫( b e^-b db).
Changing the Lim. infinity = 2.
= ∫( b e^-b db). ----------------------------(2).
The step (2) above can be solve with Integrations by part;
Lim. 2 ---> infinity [ b e^-b + ∫( 1 e^-b db|
( | = Upper boundary = 2 and the lower boundary = 0).
Lim. 2 ----> infinity [ 2 e^-2 - 0] - lim. 2 --> infinity [ 0 - 1].
Lim 2 ---> infinity [ 2/e^2 + 1].
Let 2/e^2 = j.
The, lim 2 ---> infinity j + 1.
To solve this, there is need to make use of L'hospital rule,
j = Lim 2 ---> infinity [ 1 /e^2] = 0.
Thus, j = 0. And 0 + 1 = 1.
(PROVED TO BE TRUE).
(B). Taking limit from 1 (upper) to 0( lower). Assuming that b = 3 - 4x.
Therefore, -4dx = db.
∫( 8 /(3 - 4x)^3 dx.
Taking the upper boundary = -1 and the lower boundary = 3.
∫ ( - 2db/ b^3.
= 2 ∫ ( - db/ b^3.
= 2 [ b^-3 + 2) / -3 + 1.
= 2 | - 1/ 2 db
= 8/9.
Not true.
