Question

Evaluate the definite integral

Answer 1

\[\int\limits_{1}^{2}x \sqrt{x-1} dx\]

Answer 2

can I simplify the function first?

Answer 3

try \(u = x-1\)

Answer 4

then du=dx

Answer 5

what do I do with the x?

Answer 6

\(x = u+1\)

Answer 7

put x= (sec^2)x

Answer 8

0.o

Answer 9

I get that the anitderivative of x is x^2 /2 and for u^(1/2) it's (2/3)u^(3/2)

Answer 10

\[\int (u+1) \sqrt{u} ~du= \int~ u^{\frac{3}{2}} + u^{\frac{1}{2}} du\]

Answer 11

how did you get this result?

Answer 12

ah, ok nvm I understand this part

Answer 13

Another interesting substitution, x=u^2+1.

Answer 14

so, I have \[\frac{ 2u ^{(5/2)} }{ 5 }+\frac{ 2u ^{3/2} }{ 3 } +C\]

Answer 15

If I sub in x-1, is this the final result?

Answer 16

Yes, you put now the x-1 and you will obtain a F(x). Then you should evaluate the definite integral, Integral= F(2)-F(1)

Answer 17

great :)

Answer 18

So 16/15 as final result?