definate inegrals?

Question

2 over 1 integral (x sqrt(x-1)) dx. I got the answer 2/3 but the book said it was wrong. I did u = x-1, du = dx. then 2-1 = 1 and 1=1 = 0.. then 1 over 0 integral (sqrt (u) du) then (2/3)(u^(2/3)) and finally (2/3)(1-0) = 2/3 but book got something pretty off and I think i got the u and du wrong but i dont know what is rigth then.

Sorry, but none of the comments helped me enough to get the right answer (or were wrong themselves). any other suggestins?

Answer 1

put x-1 = u; dx = du; when x = 1, u = 0; when x = 2, u = 1.
I = 1 over 0 integral [ (u + 1 ) sqrt( u ) du ]
= 1 over 0 integral [ u^ (3/2) + u^ (1/2) du ]
= 1 over 0 [ (2/5) u^ (5/2) + (2/3) u^ (3/2) ]
= 2/5 + 2/3
= 16/15

Answer 2

1

Answer 3

qwert is correct, but then:

integral(0,1){2(u^4+u^2)du} == {2(u^5)/5 + 2(u^3)/3} ==

2*1/5 + 2*1/3 == 2/5 + 2/3 == 6/15 + 10/15 == 16/15

Answer 4

put x-1 = u²

dx = 2udu

new limits are 0 to 1

we get integral (u² +1)*u*2udu with limits 0 to 1

or integral 2(u^4 +u²)du with limits 0 to 1

integrate and apply limits

Answer 5

Check your substitution. If u=x-1 then x=u+1 and your integrand becomes (U+1)sqrt(U)dU. This in turn becomes two integrals:
Usqrt(U)dU +sqrt(U)dU. You need to evaluate both of them.

Answer 6

You miss on sqrt(u)du.

Answer 7

did you use the product rule? it seems "x" and "sq rt (x-1)" are in product form. use product rule first and integrate and then substitute the upper and lower limit. if the book says wrong again, then throw that book.

Answer 8

why are you doing calculus in the beginning of august?