MATHEMATICS II - Deneme Sınavı - 18

if u = (x² + 2) and du = 2x dx then ∫u² - 4x.dx  - ∫5dx which is equal to ∫u² - 2du  - ∫5dx this yields to u³/3 - 2u - 5x. Then we can plug in u = (x² + 2) which would equal to (x²+2)³/3 - 2(x²+2) - 5x

For the interval [-1,1] we first find x=-1 which would yield to [(-1)²+2)³/3 - 2[(-1)²+2] - 5.(-1) = 27/3 - 6 +5 = 9 - 1 = 8

for x = 1  (1²+2)³/3 - 2(0²+2) - 5.0 = 27/3 - 6 - 5 = -2

F(1) - F(-1) = -2 - 8 = -10

The answer is A.

Solving the integral yields to the result of ∫x² - x + 4 dx = x³/3 - x²/2 + 4x.

For x =0, F(0) =  0³/3 - 0²/2 + 4.0 = 0

For x =2, F(2) =  2³/3 - 2²/2 + 4.2 = 8/3 - 2 + 8 = 8/3 + 6 = 26/3

F(2) - F(0) = 26/3

The answer is B.

For the function x² + 2x + 4 on the interval [0,1], a=0 and b =1. The integral ∫x² + 2x + 4 dx = x³/3+ 2x²/2 + 4x.

For x = 0, F(0)= 0³/3+ 2.0²/2 + 4.0 = 0

For x = 1, F(1)= 1³/3+ 2.1²/2 + 4.1 = 1/3 + 1 + 4 =16/3

F(1) - F(0) = 16/3 - 0 = 16/3

f_avg = 1/(1-0) . 16/3 = 16/3

The answer is E.

For x = 0, F(0)= 0³/3+ 2.0²/2 + 4.0 = 0

For x = 1, F(1)= 1³/3+ 2.1²/2 + 4.1 = 1/3 + 1 + 4 =16/3

F(1) - F(0) = 16/3 - 0 = 16/3

The answer is E.

If n=5 then the formula yields to [5.(5+1).(2.5+1)]/6 = (5.6.11)/6 =55

The answer is B.

For the function 6x² + 2x on the interval [1,3], a=1 and b =3. The integral ∫6x² + 2x dx = 6x³/3+ 2x²/2 =2x³+ x²

For x = 1, F(1)= 2.1³+ 1² = 2 + 1 = 3

For x = 3, F(3)= 2.3³+ 3² = 2.27 + 9 = 63

F(3) - F(1) = 63 - 3 = 60

f_avg = 1/(3-1) . 60 = 1/2.60 =30

The answer is D.

As also understood from the information given, the answer is E. All of the statements in the options related to the aplications of integrals in economics are correctç