How to solve this integration question?

Question

Let 𝑓(𝑥) = ∫ (1-t)/(tˆ4+2) dt which the value t is 0 to x

Determine if the following statements are true or false without explicitly computing 𝑓(𝑥). Justify all of your answers.

a) 𝑓(0) = 0

b) 𝑓(1) > 0

c) 𝑓(−1) > 0

d) 𝑓 is always increasing for 𝑥 ≥ 0

e) 𝑓 has a local maximum at 𝑥 = 1

Answer 1

y = (1-t) / (t⁴+2) Notice that the denominator is always positive. The numerator is positive when t < 1 and is negative when t > 1.

So, y > 0 if t < 1 and y < 0 if t > 1.

f(x) = ∫_{(0 to x)} [(1-t) / (t⁴+2)]dt

a. TRUE f(0) = ∫_{(0 to 0)} [(1-t) / (t⁴+2)]dt = 0. If the lower and upper limits of integration are equal, the value of the integral is zero.

b. TRUE f(1) = ∫_{(0 to 1)}[(1-t) / (t⁴+2)]dt > 0. If a < b, then the definite integral from a to b of a positive function is positive.

c. FALSE f(-1) = ∫_{(0 to -1)} [(1-t) / (t⁴+2)]dt = -∫_{(-1 to 0)}[(1-t) / (t⁴+2)]dt is not greater than zero. If a < b, then the definite integral from a to b of a positive function is positive

d. FALSE f'(x) = (1-x) / (x⁴+2) < 0 when x > 1. So, f(x) is decreasing when x > 1.

e. TRUE f'(x) = (1-x) / (x⁴+2) = 0 when x = 1. When x < 1, f'(x) > 0, so f is increasing on (-∞,1),

When x > 1, f'(x) < 0, so f is decreasing on (1,∞). Therefore, f(x) has a relative max when x = 1.