Question

help calc1 integral

Answer 1

b is easy.. 0 because you're taking the derivative of a constant (definite integral is a number)

Answer 2

\[d/dx \int\limits\limits_{g(x)}^{h(x)} f(x) dx\] = f(h(x)) h'(x) - f(g(x)) g'(x)
hope it helps..

Answer 3

can anyone help solve, my professor's notes are confusing

Answer 4

i understand the theorum for b but im not sure if im doing it right even if i got 0 for b

Answer 5

a defintite integral with constants as lower and upper limits always yield some constant ( which is understood as its an area under some curve) ..and d/dx of constant = 0 .. thus b part's ans is 0..
rest you should try yourself..

Answer 6

i know b is 0
but my professor wrote d/dx[6^(1/2)/4]=0, which is the same as the answer for a w/o taking the derivative of it

Answer 7

i know the answers, but i want to know the procedure because the answers from my professor's notes doesnt make sense, i want to konw if its right

Answer 8

why dont we do it the other way round,,you post your solutions and we'll tell whether they are correct or not.. ;)

Answer 9

ok

Answer 10

it doesnt scan right on my printer
let me try to type it out

Answer 11

a) sin[(pi/2)(1/2)]cos[(pi/2)(1/3)] - sin(0)cos(0) = sin(pi/4)cos(pi/6) -0
=[2^(1/2)/2](1/2) = [2^(1/2)/4]

Answer 12

b) = d/dx(-[2^(1/2)/2][3^(1/2)/2]-0) = -d/dx[6^(1/2)/4 = -0 =0

Answer 13

c)= -d/dx[-cos(x/2)sin(x/3)] =d/dx[cos(x/2)sin(x/3] = -sin(x/2)cos(x/3)

Answer 14

my answers are inconsistent w/ my professor's notes
because he got [6^(1/2)/4] on a), and d/dx[6^(1/2)/4]=0 on b)

Answer 15

can anyone tell me if my answers are correct or not?..... =/