Question

6
∫(x+4)√(6-2x) anyone?
2

Answer 1

u=6-2x
...

Answer 2

your limits are not good for this problem

Answer 3

i got that far... and du=-2dx... but how does that work into the eqn

Answer 4

are you sure your typed the problem correctly?

Answer 5

should have dx after √(6-2x)

Answer 6

\[\sqrt{6-2x}\] is a complex when x>3

Answer 7

* a complex number

Answer 8

are you sure that the limits of integration are from x=2 to x=6?

Answer 9

as written on my final review sheet

Answer 10

then it is a typo

Answer 11

not surprising, but could I integrate (x+4) as x^2/2+4x and √(6-2x) as (6-2x)^1/2...[(6-2x)^3/2]/3/2

Answer 12

if you just wanted to compute

\[\int(x+4)\sqrt{6-2x}dx\]

u=6-2x
du=-2dx
\[x=\frac{6-u}{2}\]

\[\int(\frac{6-u}{2}+4)\sqrt{u}\frac{du}{-2}\]

expand...then integrate term by term

Answer 13

it doesnt work as a definite integral to find area under curve of the eqn. on the interval (2,6)

Answer 14

no

Answer 15

ok thanks, a typo wouldn't be unusual