Question

AP CALCULUS!!!! Need a clear explanation and answers! Thank you!

Answer 1

Answer:

see below

Explanation:

the intervals where f(t) is positive is when the piece wise function has a positive value (not y, y is the integral of f(t))

a) f is positive over [1.4, 3.4]

   f is negative over [0, 1.4] ∪ [3.4, 6]

   f is zero over intervals {1.4} ∪ {3.4} ∪ {6}

b) maximum = 2

   minimum = -3

   I don't know why it shows that your answer is wrong, if it keeps        showing its wrong, try the integral max and min

the integral max is 2.3

the integral min is -3.1

c) avg value of integral $\int\limits^a_b {f(x)} \, dx$is: $\frac{1}{a+b} \int\limits^a_b {f(x)} \, dx$, so

avg value of $y = \int\limits^x_0 {f(t)} \, dt$  over [0, 6] is: $\frac{1}{6-0} \int\limits^6_0 {f(t)} \, dt$

$\int\limits^6_0 {f(t)} \, dt$= $\frac{(-1)(1.4)}{2} + \frac{(2+1)(2)}{2} + \frac{(2.6+1)(-3)}{2} = -0.7 + 3 -5.4 = -3.1$

$\frac{1}{6-0} \int\limits^6_0 {f(t)} \, dt$ = $\frac{1}{6} (-3.1) = \frac{-3.1}{6} = -0.516666...$ or $\frac{-31}{60}$

avg value of $y = \int\limits^x_0 {f(t)} \, dt$  over [0, 6]: $\frac{-31}{60}$