Question

1. y=∫upper bound is sqrt(x) lower bound is 1, cos(2t)/t^9 dt

using the appropriate form of the Fundamental Theorem of Calculus.
y′ =

2. Use part I of the Fundamental Theorem of Calculus to find the derivative of

F(x)=∫upper bound is 5 lower bound is x, tan(t^4)dt

F′(x) =

3. If h(x)=∫upper bound is 3/x and lower bound is 2, 9arctan t dt , then h′(x)=

4. Consider the function f(x) = {x if x<1, 1/x if x is >_ 1}

Evaluate the definite integral ∫5 is the upper bound and the lower bound is -3, f(x)dx =

Evaluate the average value of f on the interval [−3,5]

Answer 1